Find the Equation of the Parabola (-10,-28) , (-6,-21) , (-2,-14)

$(- 10, - 28)$ , $(- 6, - 21)$ , $(- 2, - 14)$

Use the standard form of a quadratic equation $y = a x^{2} + b x + c$ as the starting point for finding the equation through the three points.

$y = a x^{2} + b x + c$

Create a system of equations by substituting the $x$ and $y$ values of each point into the standard formula of a quadratic equation to create the three equation system.

$- 28 = a {(- 10)}^{2} + b (- 10) + c$

$- 21 = a {(- 6)}^{2} + b (- 6) + c$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

Solve the system of equations to find the values of $a$ , $b$ , and $c$ .

Solve for $c$ in $- 28 = a {(- 10)}^{2} + b (- 10) + c$ .

Rewrite the equation as $a {(- 10)}^{2} + b (- 10) + c = - 28$ .

$a {(- 10)}^{2} + b (- 10) + c = - 28$

$- 21 = a {(- 6)}^{2} + b (- 6) + c$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

Simplify each term.

Raise $- 10$ to the power of $2$ .

$a \cdot 100 + b (- 10) + c = - 28$

$- 21 = a {(- 6)}^{2} + b (- 6) + c$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

Move $100$ to the left of $a$ .

$100 \cdot a + b (- 10) + c = - 28$

$- 21 = a {(- 6)}^{2} + b (- 6) + c$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

Move $- 10$ to the left of $b$ .

$100 a - 10 b + c = - 28$

$- 21 = a {(- 6)}^{2} + b (- 6) + c$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

$100 a - 10 b + c = - 28$

$- 21 = a {(- 6)}^{2} + b (- 6) + c$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

Move all terms not containing $c$ to the right side of the equation.

Subtract $100 a$ from both sides of the equation.

$- 10 b + c = - 28 - 100 a$

$- 21 = a {(- 6)}^{2} + b (- 6) + c$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

Add $10 b$ to both sides of the equation.

$c = - 28 - 100 a + 10 b$

$- 21 = a {(- 6)}^{2} + b (- 6) + c$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

$c = - 28 - 100 a + 10 b$

$- 21 = a {(- 6)}^{2} + b (- 6) + c$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

$c = - 28 - 100 a + 10 b$

$- 21 = a {(- 6)}^{2} + b (- 6) + c$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

Replace all occurrences of $c$ with $- 28 - 100 a + 10 b$ in each equation.

Replace all occurrences of $c$ in $- 21 = a {(- 6)}^{2} + b (- 6) + c$ with $- 28 - 100 a + 10 b$ .

$- 21 = a {(- 6)}^{2} + b (- 6) - 28 - 100 a + 10 b$

$c = - 28 - 100 a + 10 b$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

Simplify $- 21 = a {(- 6)}^{2} + b (- 6) - 28 - 100 a + 10 b$ .

Simplify the left side.

Remove parentheses.

$- 21 = a {(- 6)}^{2} + b (- 6) - 28 - 100 a + 10 b$

$c = - 28 - 100 a + 10 b$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

$- 21 = a {(- 6)}^{2} + b (- 6) - 28 - 100 a + 10 b$

$c = - 28 - 100 a + 10 b$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

Simplify the right side.

Simplify $a {(- 6)}^{2} + b (- 6) - 28 - 100 a + 10 b$ .

Simplify each term.

Raise $- 6$ to the power of $2$ .

$- 21 = a \cdot 36 + b (- 6) - 28 - 100 a + 10 b$

$c = - 28 - 100 a + 10 b$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

Move $36$ to the left of $a$ .

$- 21 = 36 \cdot a + b (- 6) - 28 - 100 a + 10 b$

$c = - 28 - 100 a + 10 b$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

Move $- 6$ to the left of $b$ .

$- 21 = 36 a - 6 b - 28 - 100 a + 10 b$

$c = - 28 - 100 a + 10 b$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

$- 21 = 36 a - 6 b - 28 - 100 a + 10 b$

$c = - 28 - 100 a + 10 b$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

Simplify by adding terms.

Subtract $100 a$ from $36 a$ .

$- 21 = - 64 a - 6 b - 28 + 10 b$

$c = - 28 - 100 a + 10 b$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

Add $- 6 b$ and $10 b$ .

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$- 14 = a {(- 2)}^{2} + b (- 2) + c$

Replace all occurrences of $c$ in $- 14 = a {(- 2)}^{2} + b (- 2) + c$ with $- 28 - 100 a + 10 b$ .

$- 14 = a {(- 2)}^{2} + b (- 2) - 28 - 100 a + 10 b$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Simplify $- 14 = a {(- 2)}^{2} + b (- 2) - 28 - 100 a + 10 b$ .

Simplify the left side.

Remove parentheses.

$- 14 = a {(- 2)}^{2} + b (- 2) - 28 - 100 a + 10 b$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$- 14 = a {(- 2)}^{2} + b (- 2) - 28 - 100 a + 10 b$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Simplify the right side.

Simplify $a {(- 2)}^{2} + b (- 2) - 28 - 100 a + 10 b$ .

Simplify each term.

Raise $- 2$ to the power of $2$ .

$- 14 = a \cdot 4 + b (- 2) - 28 - 100 a + 10 b$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Move $4$ to the left of $a$ .

$- 14 = 4 \cdot a + b (- 2) - 28 - 100 a + 10 b$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Move $- 2$ to the left of $b$ .

$- 14 = 4 a - 2 b - 28 - 100 a + 10 b$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$- 14 = 4 a - 2 b - 28 - 100 a + 10 b$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Simplify by adding terms.

Subtract $100 a$ from $4 a$ .

$- 14 = - 96 a - 2 b - 28 + 10 b$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Add $- 2 b$ and $10 b$ .

$- 14 = - 96 a + 8 b - 28$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$- 14 = - 96 a + 8 b - 28$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$- 14 = - 96 a + 8 b - 28$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$- 14 = - 96 a + 8 b - 28$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$- 14 = - 96 a + 8 b - 28$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$- 14 = - 96 a + 8 b - 28$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Solve for $a$ in $- 14 = - 96 a + 8 b - 28$ .

Rewrite the equation as $- 96 a + 8 b - 28 = - 14$ .

$- 96 a + 8 b - 28 = - 14$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Move all terms not containing $a$ to the right side of the equation.

Subtract $8 b$ from both sides of the equation.

$- 96 a - 28 = - 14 - 8 b$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Add $28$ to both sides of the equation.

$- 96 a = - 14 - 8 b + 28$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Add $- 14$ and $28$ .

$- 96 a = - 8 b + 14$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$- 96 a = - 8 b + 14$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Divide each term in $- 96 a = - 8 b + 14$ by $- 96$ and simplify.

Divide each term in $- 96 a = - 8 b + 14$ by $- 96$ .

$\frac{- 96 a}{- 96} = \frac{- 8 b}{- 96} + \frac{14}{- 96}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Simplify the left side.

Cancel the common factor of $- 96$ .

Cancel the common factor.

$\frac{- 96 a}{- 96} = \frac{- 8 b}{- 96} + \frac{14}{- 96}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Divide $a$ by $1$ .

$a = \frac{- 8 b}{- 96} + \frac{14}{- 96}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$a = \frac{- 8 b}{- 96} + \frac{14}{- 96}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$a = \frac{- 8 b}{- 96} + \frac{14}{- 96}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 8$ and $- 96$ .

Factor $- 8$ out of $- 8 b$ .

$a = \frac{- 8 b}{- 96} + \frac{14}{- 96}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Cancel the common factors.

Factor $- 8$ out of $- 96$ .

$a = \frac{- 8 b}{- 8 \cdot 12} + \frac{14}{- 96}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Cancel the common factor.

$a = \frac{- 8 b}{- 8 \cdot 12} + \frac{14}{- 96}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Rewrite the expression.

$a = \frac{b}{12} + \frac{14}{- 96}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$a = \frac{b}{12} + \frac{14}{- 96}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$a = \frac{b}{12} + \frac{14}{- 96}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Cancel the common factor of $14$ and $- 96$ .

Factor $2$ out of $14$ .

$a = \frac{b}{12} + \frac{2 (7)}{- 96}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Cancel the common factors.

Factor $2$ out of $- 96$ .

$a = \frac{b}{12} + \frac{2 \cdot 7}{2 \cdot - 48}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Cancel the common factor.

$a = \frac{b}{12} + \frac{2 \cdot 7}{2 \cdot - 48}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Rewrite the expression.

$a = \frac{b}{12} + \frac{7}{- 48}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$a = \frac{b}{12} + \frac{7}{- 48}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$a = \frac{b}{12} + \frac{7}{- 48}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Move the negative in front of the fraction.

$a = \frac{b}{12} - \frac{7}{48}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$a = \frac{b}{12} - \frac{7}{48}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$a = \frac{b}{12} - \frac{7}{48}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$a = \frac{b}{12} - \frac{7}{48}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

$a = \frac{b}{12} - \frac{7}{48}$

$- 21 = - 64 a + 4 b - 28$

$c = - 28 - 100 a + 10 b$

Replace all occurrences of $a$ with $\frac{b}{12} - \frac{7}{48}$ in each equation.

Replace all occurrences of $a$ in $- 21 = - 64 a + 4 b - 28$ with $\frac{b}{12} - \frac{7}{48}$ .

$- 21 = - 64 (\frac{b}{12} - \frac{7}{48}) + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Simplify the right side.

Simplify $- 64 (\frac{b}{12} - \frac{7}{48}) + 4 b - 28$ .

Simplify each term.

Apply the distributive property.

$- 21 = - 64 \frac{b}{12} - 64 (- \frac{7}{48}) + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Cancel the common factor of $4$ .

Factor $4$ out of $- 64$ .

$- 21 = 4 (- 16) (\frac{b}{12}) - 64 (- \frac{7}{48}) + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Factor $4$ out of $12$ .

$- 21 = 4 \cdot (- 16 \frac{b}{4 \cdot 3}) - 64 (- \frac{7}{48}) + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Cancel the common factor.

$- 21 = 4 \cdot (- 16 \frac{b}{4 \cdot 3}) - 64 (- \frac{7}{48}) + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Rewrite the expression.

$- 21 = - 16 \frac{b}{3} - 64 (- \frac{7}{48}) + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

$- 21 = - 16 \frac{b}{3} - 64 (- \frac{7}{48}) + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Combine $- 16$ and $\frac{b}{3}$ .

$- 21 = \frac{- 16 b}{3} - 64 (- \frac{7}{48}) + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Cancel the common factor of $16$ .

Move the leading negative in $- \frac{7}{48}$ into the numerator.

$- 21 = \frac{- 16 b}{3} - 64 (\frac{- 7}{48}) + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Factor $16$ out of $- 64$ .

$- 21 = \frac{- 16 b}{3} + 16 (- 4) (\frac{- 7}{48}) + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Factor $16$ out of $48$ .

$- 21 = \frac{- 16 b}{3} + 16 \cdot (- 4 \frac{- 7}{16 \cdot 3}) + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Cancel the common factor.

$- 21 = \frac{- 16 b}{3} + 16 \cdot (- 4 \frac{- 7}{16 \cdot 3}) + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Rewrite the expression.

$- 21 = \frac{- 16 b}{3} - 4 (\frac{- 7}{3}) + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

$- 21 = \frac{- 16 b}{3} - 4 (\frac{- 7}{3}) + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Combine $- 4$ and $\frac{- 7}{3}$ .

$- 21 = \frac{- 16 b}{3} + \frac{- 4 \cdot - 7}{3} + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Multiply $- 4$ by $- 7$ .

$- 21 = \frac{- 16 b}{3} + \frac{28}{3} + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Move the negative in front of the fraction.

$- 21 = - \frac{16 b}{3} + \frac{28}{3} + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

$- 21 = - \frac{16 b}{3} + \frac{28}{3} + 4 b - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

To write $4 b$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- 21 = - \frac{16 b}{3} + 4 b \cdot \frac{3}{3} + \frac{28}{3} - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Combine $4 b$ and $\frac{3}{3}$ .

$- 21 = - \frac{16 b}{3} + \frac{4 b \cdot 3}{3} + \frac{28}{3} - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Combine the numerators over the common denominator.

$- 21 = \frac{- 16 b + 4 b \cdot 3}{3} + \frac{28}{3} - 28$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Combine the numerators over the common denominator.

$- 21 = - 28 + \frac{- 16 b + 4 b \cdot 3 + 28}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Multiply $3$ by $4$ .

$- 21 = - 28 + \frac{- 16 b + 12 b + 28}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Add $- 16 b$ and $12 b$ .

$- 21 = - 28 + \frac{- 4 b + 28}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Factor $4$ out of $- 4 b + 28$ .

Factor $4$ out of $- 4 b$ .

$- 21 = - 28 + \frac{4 (- b) + 28}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Factor $4$ out of $28$ .

$- 21 = - 28 + \frac{4 (- b) + 4 (7)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Factor $4$ out of $4 (- b) + 4 (7)$ .

$- 21 = - 28 + \frac{4 (- b + 7)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

$- 21 = - 28 + \frac{4 (- b + 7)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

To write $- 28$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- 21 = - 28 \cdot \frac{3}{3} + \frac{4 (- b + 7)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Simplify terms.

Combine $- 28$ and $\frac{3}{3}$ .

$- 21 = \frac{- 28 \cdot 3}{3} + \frac{4 (- b + 7)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Combine the numerators over the common denominator.

$- 21 = \frac{- 28 \cdot 3 + 4 (- b + 7)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

$- 21 = \frac{- 28 \cdot 3 + 4 (- b + 7)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Simplify the numerator.

Factor $4$ out of $- 28 \cdot 3 + 4 (- b + 7)$ .

Factor $4$ out of $- 28 \cdot 3$ .

$- 21 = \frac{4 (- 7 \cdot 3) + 4 (- b + 7)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Factor $4$ out of $4 (- 7 \cdot 3) + 4 (- b + 7)$ .

$- 21 = \frac{4 (- 7 \cdot 3 - b + 7)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

$- 21 = \frac{4 (- 7 \cdot 3 - b + 7)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Multiply $- 7$ by $3$ .

$- 21 = \frac{4 (- 21 - b + 7)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Add $- 21$ and $7$ .

$- 21 = \frac{4 (- b - 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

$- 21 = \frac{4 (- b - 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Simplify with factoring out.

Factor $- 1$ out of $- b$ .

$- 21 = \frac{4 (- (b) - 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Rewrite $- 14$ as $- 1 (14)$ .

$- 21 = \frac{4 (- (b) - 1 \cdot 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Factor $- 1$ out of $- (b) - 1 (14)$ .

$- 21 = \frac{4 (- (b + 14))}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Simplify the expression.

Rewrite $- (b + 14)$ as $- 1 (b + 14)$ .

$- 21 = \frac{4 (- 1 (b + 14))}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Move the negative in front of the fraction.

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 100 a + 10 b$

Replace all occurrences of $a$ in $c = - 28 - 100 a + 10 b$ with $\frac{b}{12} - \frac{7}{48}$ .

$c = - 28 - 100 (\frac{b}{12} - \frac{7}{48}) + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Simplify the right side.

Simplify $- 28 - 100 (\frac{b}{12} - \frac{7}{48}) + 10 b$ .

Simplify each term.

Apply the distributive property.

$c = - 28 - 100 \frac{b}{12} - 100 (- \frac{7}{48}) + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Cancel the common factor of $4$ .

Factor $4$ out of $- 100$ .

$c = - 28 + 4 (- 25) (\frac{b}{12}) - 100 (- \frac{7}{48}) + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Factor $4$ out of $12$ .

$c = - 28 + 4 \cdot (- 25 \frac{b}{4 \cdot 3}) - 100 (- \frac{7}{48}) + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Cancel the common factor.

$c = - 28 + 4 \cdot (- 25 \frac{b}{4 \cdot 3}) - 100 (- \frac{7}{48}) + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Rewrite the expression.

$c = - 28 - 25 \frac{b}{3} - 100 (- \frac{7}{48}) + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - 25 \frac{b}{3} - 100 (- \frac{7}{48}) + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Combine $- 25$ and $\frac{b}{3}$ .

$c = - 28 + \frac{- 25 b}{3} - 100 (- \frac{7}{48}) + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{7}{48}$ into the numerator.

$c = - 28 + \frac{- 25 b}{3} - 100 (\frac{- 7}{48}) + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Factor $4$ out of $- 100$ .

$c = - 28 + \frac{- 25 b}{3} + 4 (- 25) (\frac{- 7}{48}) + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Factor $4$ out of $48$ .

$c = - 28 + \frac{- 25 b}{3} + 4 \cdot (- 25 \frac{- 7}{4 \cdot 12}) + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Cancel the common factor.

$c = - 28 + \frac{- 25 b}{3} + 4 \cdot (- 25 \frac{- 7}{4 \cdot 12}) + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Rewrite the expression.

$c = - 28 + \frac{- 25 b}{3} - 25 (\frac{- 7}{12}) + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 + \frac{- 25 b}{3} - 25 (\frac{- 7}{12}) + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Combine $- 25$ and $\frac{- 7}{12}$ .

$c = - 28 + \frac{- 25 b}{3} + \frac{- 25 \cdot - 7}{12} + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Multiply $- 25$ by $- 7$ .

$c = - 28 + \frac{- 25 b}{3} + \frac{175}{12} + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Move the negative in front of the fraction.

$c = - 28 - \frac{25 b}{3} + \frac{175}{12} + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - 28 - \frac{25 b}{3} + \frac{175}{12} + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

To write $- 28$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

$c = - \frac{25 b}{3} - 28 \cdot \frac{12}{12} + \frac{175}{12} + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Combine $- 28$ and $\frac{12}{12}$ .

$c = - \frac{25 b}{3} + \frac{- 28 \cdot 12}{12} + \frac{175}{12} + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Combine the numerators over the common denominator.

$c = - \frac{25 b}{3} + \frac{- 28 \cdot 12 + 175}{12} + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Simplify the numerator.

Multiply $- 28$ by $12$ .

$c = - \frac{25 b}{3} + \frac{- 336 + 175}{12} + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Add $- 336$ and $175$ .

$c = - \frac{25 b}{3} + \frac{- 161}{12} + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - \frac{25 b}{3} + \frac{- 161}{12} + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Move the negative in front of the fraction.

$c = - \frac{25 b}{3} - \frac{161}{12} + 10 b$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

To write $10 b$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$c = - \frac{25 b}{3} + 10 b \cdot \frac{3}{3} - \frac{161}{12}$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Simplify terms.

Combine $10 b$ and $\frac{3}{3}$ .

$c = - \frac{25 b}{3} + \frac{10 b \cdot 3}{3} - \frac{161}{12}$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Combine the numerators over the common denominator.

$c = \frac{- 25 b + 10 b \cdot 3}{3} - \frac{161}{12}$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = \frac{- 25 b + 10 b \cdot 3}{3} - \frac{161}{12}$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Simplify the numerator.

Factor $5 b$ out of $- 25 b + 10 b \cdot 3$ .

Factor $5 b$ out of $- 25 b$ .

$c = \frac{5 b (- 5) + 10 b \cdot 3}{3} - \frac{161}{12}$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Factor $5 b$ out of $10 b \cdot 3$ .

$c = \frac{5 b (- 5) + 5 b (2 \cdot 3)}{3} - \frac{161}{12}$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Factor $5 b$ out of $5 b (- 5) + 5 b (2 \cdot 3)$ .

$c = \frac{5 b (- 5 + 2 \cdot 3)}{3} - \frac{161}{12}$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = \frac{5 b (- 5 + 2 \cdot 3)}{3} - \frac{161}{12}$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Multiply $2$ by $3$ .

$c = \frac{5 b (- 5 + 6)}{3} - \frac{161}{12}$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Add $- 5$ and $6$ .

$c = \frac{5 b \cdot 1}{3} - \frac{161}{12}$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Multiply $5$ by $1$ .

$c = \frac{5 b}{3} - \frac{161}{12}$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$- 21 = - \frac{4 (b + 14)}{3}$

$a = \frac{b}{12} - \frac{7}{48}$

Solve for $b$ in $- 21 = - \frac{4 (b + 14)}{3}$ .

Rewrite the equation as $- \frac{4 (b + 14)}{3} = - 21$ .

$- \frac{4 (b + 14)}{3} = - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Multiply both sides of the equation by $- \frac{3}{4}$ .

$- \frac{3}{4} \cdot (- \frac{4 (b + 14)}{3}) = - \frac{3}{4} \cdot - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- \frac{3}{4} (- \frac{4 (b + 14)}{3})$ .

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{3}{4}$ into the numerator.

$\frac{- 3}{4} \cdot (- \frac{4 (b + 14)}{3}) = - \frac{3}{4} \cdot - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Move the leading negative in $- \frac{4 (b + 14)}{3}$ into the numerator.

$\frac{- 3}{4} \cdot \frac{- 4 (b + 14)}{3} = - \frac{3}{4} \cdot - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Factor $3$ out of $- 3$ .

$\frac{3 (- 1)}{4} \cdot \frac{- 4 (b + 14)}{3} = - \frac{3}{4} \cdot - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Cancel the common factor.

$\frac{3 \cdot - 1}{4} \cdot \frac{- 4 (b + 14)}{3} = - \frac{3}{4} \cdot - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Rewrite the expression.

$\frac{- 1}{4} \cdot (- 4 (b + 14)) = - \frac{3}{4} \cdot - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

$\frac{- 1}{4} \cdot (- 4 (b + 14)) = - \frac{3}{4} \cdot - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Cancel the common factor of $4$ .

Factor $4$ out of $- 4 (b + 14)$ .

$\frac{- 1}{4} \cdot (4 (- (b + 14))) = - \frac{3}{4} \cdot - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Cancel the common factor.

$\frac{- 1}{4} \cdot (4 (- (b + 14))) = - \frac{3}{4} \cdot - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Rewrite the expression.

$b + 14 = - \frac{3}{4} \cdot - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

$b + 14 = - \frac{3}{4} \cdot - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 (b + 14) = - \frac{3}{4} \cdot - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Multiply $b + 14$ by $1$ .

$b + 14 = - \frac{3}{4} \cdot - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

$b + 14 = - \frac{3}{4} \cdot - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

$b + 14 = - \frac{3}{4} \cdot - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

$b + 14 = - \frac{3}{4} \cdot - 21$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Simplify the right side.

Multiply $- \frac{3}{4} \cdot - 21$ .

Multiply $- 21$ by $- 1$ .

$b + 14 = 21 (\frac{3}{4})$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Combine $21$ and $\frac{3}{4}$ .

$b + 14 = \frac{21 \cdot 3}{4}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Multiply $21$ by $3$ .

$b + 14 = \frac{63}{4}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

$b + 14 = \frac{63}{4}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

$b + 14 = \frac{63}{4}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

$b + 14 = \frac{63}{4}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Move all terms not containing $b$ to the right side of the equation.

Subtract $14$ from both sides of the equation.

$b = \frac{63}{4} - 14$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

To write $- 14$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$b = \frac{63}{4} - 14 \cdot \frac{4}{4}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Combine $- 14$ and $\frac{4}{4}$ .

$b = \frac{63}{4} + \frac{- 14 \cdot 4}{4}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Combine the numerators over the common denominator.

$b = \frac{63 - 14 \cdot 4}{4}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Simplify the numerator.

Multiply $- 14$ by $4$ .

$b = \frac{63 - 56}{4}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Subtract $56$ from $63$ .

$b = \frac{7}{4}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

$b = \frac{7}{4}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

$b = \frac{7}{4}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

$b = \frac{7}{4}$

$c = \frac{5 b}{3} - \frac{161}{12}$

$a = \frac{b}{12} - \frac{7}{48}$

Replace all occurrences of $b$ with $\frac{7}{4}$ in each equation.

Replace all occurrences of $b$ in $c = \frac{5 b}{3} - \frac{161}{12}$ with $\frac{7}{4}$ .

$c = \frac{5 (\frac{7}{4})}{3} - \frac{161}{12}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

Simplify the right side.

Simplify $\frac{5 (\frac{7}{4})}{3} - \frac{161}{12}$ .

Simplify each term.

Combine $5$ and $\frac{7}{4}$ .

$c = \frac{\frac{5 \cdot 7}{4}}{3} - \frac{161}{12}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

Multiply $5$ by $7$ .

$c = \frac{\frac{35}{4}}{3} - \frac{161}{12}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

Multiply the numerator by the reciprocal of the denominator.

$c = \frac{35}{4} \cdot \frac{1}{3} - \frac{161}{12}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

Multiply $\frac{35}{4} \cdot \frac{1}{3}$ .

Multiply $\frac{35}{4}$ by $\frac{1}{3}$ .

$c = \frac{35}{4 \cdot 3} - \frac{161}{12}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

Multiply $4$ by $3$ .

$c = \frac{35}{12} - \frac{161}{12}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = \frac{35}{12} - \frac{161}{12}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = \frac{35}{12} - \frac{161}{12}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

Simplify terms.

Combine the numerators over the common denominator.

$c = \frac{35 - 161}{12}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

Subtract $161$ from $35$ .

$c = \frac{- 126}{12}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

Cancel the common factor of $- 126$ and $12$ .

Factor $6$ out of $- 126$ .

$c = \frac{6 (- 21)}{12}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

Cancel the common factors.

Factor $6$ out of $12$ .

$c = \frac{6 \cdot - 21}{6 \cdot 2}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

Cancel the common factor.

$c = \frac{6 \cdot - 21}{6 \cdot 2}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

Rewrite the expression.

$c = \frac{- 21}{2}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = \frac{- 21}{2}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = \frac{- 21}{2}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

Move the negative in front of the fraction.

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

$a = \frac{b}{12} - \frac{7}{48}$

Replace all occurrences of $b$ in $a = \frac{b}{12} - \frac{7}{48}$ with $\frac{7}{4}$ .

$a = \frac{\frac{7}{4}}{12} - \frac{7}{48}$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

Simplify the right side.

Simplify $\frac{\frac{7}{4}}{12} - \frac{7}{48}$ .

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$a = \frac{7}{4} \cdot \frac{1}{12} - \frac{7}{48}$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

Multiply $\frac{7}{4} \cdot \frac{1}{12}$ .

Multiply $\frac{7}{4}$ by $\frac{1}{12}$ .

$a = \frac{7}{4 \cdot 12} - \frac{7}{48}$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

Multiply $4$ by $12$ .

$a = \frac{7}{48} - \frac{7}{48}$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

$a = \frac{7}{48} - \frac{7}{48}$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

$a = \frac{7}{48} - \frac{7}{48}$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

Combine fractions.

Combine the numerators over the common denominator.

$a = \frac{7 - 7}{48}$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

Simplify the expression.

Subtract $7$ from $7$ .

$a = \frac{0}{48}$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

Divide $0$ by $48$ .

$a = 0$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

$a = 0$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

$a = 0$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

$a = 0$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

$a = 0$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

$a = 0$

$c = - \frac{21}{2}$

$b = \frac{7}{4}$

List all of the solutions.

$a = 0, c = - \frac{21}{2}, b = \frac{7}{4}$

Substitute the actual values of $a$ , $b$ , and $c$ into the formula for a quadratic equation to find the resulting equation.

$y = \frac{7 x}{4} - \frac{21}{2}$

Find the Equation of the Parabola (-10,-28) , (-6,-21) , (-2,-14)

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