Find the Equation of the Parabola (1,10) , (2,50) , (3,250)

$(1, 10)$ , $(2, 50)$ , $(3, 250)$

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.

$10 = a {(1)}^{2} + b (1) + c$

$50 = a {(2)}^{2} + b (2) + c$

$250 = a {(3)}^{2} + b (3) + c$

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

Solve for $a$ in $10 = a + b + c$ .

Rewrite the equation as $a + b + c = 10$ .

$a + b + c = 10$

$50 = a \cdot 2^{2} + b (2) + c$

$250 = a \cdot 3^{2} + b (3) + c$

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

Subtract $b$ from both sides of the equation.

$a + c = 10 - b$

$50 = a \cdot 2^{2} + b (2) + c$

$250 = a \cdot 3^{2} + b (3) + c$

Subtract $c$ from both sides of the equation.

$a = 10 - b - c$

$50 = a \cdot 2^{2} + b (2) + c$

$250 = a \cdot 3^{2} + b (3) + c$

$a = 10 - b - c$

$50 = a \cdot 2^{2} + b (2) + c$

$250 = a \cdot 3^{2} + b (3) + c$

$a = 10 - b - c$

$50 = a \cdot 2^{2} + b (2) + c$

$250 = a \cdot 3^{2} + b (3) + c$

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

Replace all occurrences of $a$ in $50 = a \cdot 2^{2} + b (2) + c$ with $10 - b - c$ .

$50 = (10 - b - c) \cdot 2^{2} + b (2) + c$

$a = 10 - b - c$

$250 = a \cdot 3^{2} + b (3) + c$

Simplify the right side.

Simplify $(10 - b - c) \cdot 2^{2} + b (2) + c$ .

Simplify each term.

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

$50 = (10 - b - c) \cdot 4 + b (2) + c$

$a = 10 - b - c$

$250 = a \cdot 3^{2} + b (3) + c$

Apply the distributive property.

$50 = 10 \cdot 4 - b \cdot 4 - c \cdot 4 + b (2) + c$

$a = 10 - b - c$

$250 = a \cdot 3^{2} + b (3) + c$

Simplify.

Multiply $10$ by $4$ .

$50 = 40 - b \cdot 4 - c \cdot 4 + b (2) + c$

$a = 10 - b - c$

$250 = a \cdot 3^{2} + b (3) + c$

Multiply $4$ by $- 1$ .

$50 = 40 - 4 b - c \cdot 4 + b (2) + c$

$a = 10 - b - c$

$250 = a \cdot 3^{2} + b (3) + c$

Multiply $4$ by $- 1$ .

$50 = 40 - 4 b - 4 c + b (2) + c$

$a = 10 - b - c$

$250 = a \cdot 3^{2} + b (3) + c$

$50 = 40 - 4 b - 4 c + b (2) + c$

$a = 10 - b - c$

$250 = a \cdot 3^{2} + b (3) + c$

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

$50 = 40 - 4 b - 4 c + 2 b + c$

$a = 10 - b - c$

$250 = a \cdot 3^{2} + b (3) + c$

$50 = 40 - 4 b - 4 c + 2 b + c$

$a = 10 - b - c$

$250 = a \cdot 3^{2} + b (3) + c$

Simplify by adding terms.

Add $- 4 b$ and $2 b$ .

$50 = 40 - 2 b - 4 c + c$

$a = 10 - b - c$

$250 = a \cdot 3^{2} + b (3) + c$

Add $- 4 c$ and $c$ .

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$250 = a \cdot 3^{2} + b (3) + c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$250 = a \cdot 3^{2} + b (3) + c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$250 = a \cdot 3^{2} + b (3) + c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$250 = a \cdot 3^{2} + b (3) + c$

Replace all occurrences of $a$ in $250 = a \cdot 3^{2} + b (3) + c$ with $10 - b - c$ .

$250 = (10 - b - c) \cdot 3^{2} + b (3) + c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Simplify the right side.

Simplify $(10 - b - c) \cdot 3^{2} + b (3) + c$ .

Simplify each term.

Raise $3$ to the power of $2$ .

$250 = (10 - b - c) \cdot 9 + b (3) + c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Apply the distributive property.

$250 = 10 \cdot 9 - b \cdot 9 - c \cdot 9 + b (3) + c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Simplify.

Multiply $10$ by $9$ .

$250 = 90 - b \cdot 9 - c \cdot 9 + b (3) + c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Multiply $9$ by $- 1$ .

$250 = 90 - 9 b - c \cdot 9 + b (3) + c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Multiply $9$ by $- 1$ .

$250 = 90 - 9 b - 9 c + b (3) + c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$250 = 90 - 9 b - 9 c + b (3) + c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Move $3$ to the left of $b$ .

$250 = 90 - 9 b - 9 c + 3 b + c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$250 = 90 - 9 b - 9 c + 3 b + c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Simplify by adding terms.

Add $- 9 b$ and $3 b$ .

$250 = 90 - 6 b - 9 c + c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Add $- 9 c$ and $c$ .

$250 = 90 - 6 b - 8 c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$250 = 90 - 6 b - 8 c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$250 = 90 - 6 b - 8 c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$250 = 90 - 6 b - 8 c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$250 = 90 - 6 b - 8 c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Solve for $b$ in $250 = 90 - 6 b - 8 c$ .

Rewrite the equation as $90 - 6 b - 8 c = 250$ .

$90 - 6 b - 8 c = 250$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

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

Subtract $90$ from both sides of the equation.

$- 6 b - 8 c = 250 - 90$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Add $8 c$ to both sides of the equation.

$- 6 b = 250 - 90 + 8 c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Subtract $90$ from $250$ .

$- 6 b = 160 + 8 c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$- 6 b = 160 + 8 c$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Divide each term in $- 6 b = 160 + 8 c$ by $- 6$ and simplify.

Divide each term in $- 6 b = 160 + 8 c$ by $- 6$ .

$\frac{- 6 b}{- 6} = \frac{160}{- 6} + \frac{8 c}{- 6}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 b}{- 6} = \frac{160}{- 6} + \frac{8 c}{- 6}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Divide $b$ by $1$ .

$b = \frac{160}{- 6} + \frac{8 c}{- 6}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$b = \frac{160}{- 6} + \frac{8 c}{- 6}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$b = \frac{160}{- 6} + \frac{8 c}{- 6}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Simplify the right side.

Simplify each term.

Cancel the common factor of $160$ and $- 6$ .

Factor $2$ out of $160$ .

$b = \frac{2 (80)}{- 6} + \frac{8 c}{- 6}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Cancel the common factors.

Factor $2$ out of $- 6$ .

$b = \frac{2 \cdot 80}{2 \cdot - 3} + \frac{8 c}{- 6}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Cancel the common factor.

$b = \frac{2 \cdot 80}{2 \cdot - 3} + \frac{8 c}{- 6}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Rewrite the expression.

$b = \frac{80}{- 3} + \frac{8 c}{- 6}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$b = \frac{80}{- 3} + \frac{8 c}{- 6}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$b = \frac{80}{- 3} + \frac{8 c}{- 6}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Move the negative in front of the fraction.

$b = - \frac{80}{3} + \frac{8 c}{- 6}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Cancel the common factor of $8$ and $- 6$ .

Factor $2$ out of $8 c$ .

$b = - \frac{80}{3} + \frac{2 (4 c)}{- 6}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Cancel the common factors.

Factor $2$ out of $- 6$ .

$b = - \frac{80}{3} + \frac{2 (4 c)}{2 (- 3)}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Cancel the common factor.

$b = - \frac{80}{3} + \frac{2 (4 c)}{2 \cdot - 3}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Rewrite the expression.

$b = - \frac{80}{3} + \frac{4 c}{- 3}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$b = - \frac{80}{3} + \frac{4 c}{- 3}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$b = - \frac{80}{3} + \frac{4 c}{- 3}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Move the negative in front of the fraction.

$b = - \frac{80}{3} - \frac{4 c}{3}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$50 = 40 - 2 b - 3 c$

$a = 10 - b - c$

Replace all occurrences of $b$ with $- \frac{80}{3} - \frac{4 c}{3}$ in each equation.

Replace all occurrences of $b$ in $50 = 40 - 2 b - 3 c$ with $- \frac{80}{3} - \frac{4 c}{3}$ .

$50 = 40 - 2 (- \frac{80}{3} - \frac{4 c}{3}) - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

Simplify the right side.

Simplify $40 - 2 (- \frac{80}{3} - \frac{4 c}{3}) - 3 c$ .

Simplify each term.

Apply the distributive property.

$50 = 40 - 2 (- \frac{80}{3}) - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

Multiply $- 2 (- \frac{80}{3})$ .

Multiply $- 1$ by $- 2$ .

$50 = 40 + 2 (\frac{80}{3}) - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

Combine $2$ and $\frac{80}{3}$ .

$50 = 40 + \frac{2 \cdot 80}{3} - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

Multiply $2$ by $80$ .

$50 = 40 + \frac{160}{3} - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

$50 = 40 + \frac{160}{3} - 2 (- \frac{4 c}{3}) - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

Multiply $- 2 (- \frac{4 c}{3})$ .

Multiply $- 1$ by $- 2$ .

$50 = 40 + \frac{160}{3} + 2 (\frac{4 c}{3}) - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

Combine $2$ and $\frac{4 c}{3}$ .

$50 = 40 + \frac{160}{3} + \frac{2 (4 c)}{3} - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

Multiply $4$ by $2$ .

$50 = 40 + \frac{160}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

$50 = 40 + \frac{160}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

$50 = 40 + \frac{160}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

To write $40$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$50 = 40 \cdot \frac{3}{3} + \frac{160}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

Combine $40$ and $\frac{3}{3}$ .

$50 = \frac{40 \cdot 3}{3} + \frac{160}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

Combine the numerators over the common denominator.

$50 = \frac{40 \cdot 3 + 160}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

Simplify the numerator.

Multiply $40$ by $3$ .

$50 = \frac{120 + 160}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

Add $120$ and $160$ .

$50 = \frac{280}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

$50 = \frac{280}{3} + \frac{8 c}{3} - 3 c$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

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

$50 = \frac{280}{3} + \frac{8 c}{3} - 3 c \cdot \frac{3}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

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

$50 = \frac{280}{3} + \frac{8 c}{3} + \frac{- 3 c \cdot 3}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

Combine the numerators over the common denominator.

$50 = \frac{280}{3} + \frac{8 c - 3 c \cdot 3}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

Combine the numerators over the common denominator.

$50 = \frac{280 + 8 c - 3 c \cdot 3}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

Multiply $3$ by $- 3$ .

$50 = \frac{280 + 8 c - 9 c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

Subtract $9 c$ from $8 c$ .

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 - b - c$

Replace all occurrences of $b$ in $a = 10 - b - c$ with $- \frac{80}{3} - \frac{4 c}{3}$ .

$a = 10 - (- \frac{80}{3} - \frac{4 c}{3}) - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Simplify the right side.

Simplify $10 - (- \frac{80}{3} - \frac{4 c}{3}) - c$ .

Simplify each term.

Apply the distributive property.

$a = 10 + \frac{80}{3} + \frac{4 c}{3} - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Multiply $- - \frac{80}{3}$ .

Multiply $- 1$ by $- 1$ .

$a = 10 + 1 (\frac{80}{3}) + \frac{4 c}{3} - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Multiply $\frac{80}{3}$ by $1$ .

$a = 10 + \frac{80}{3} + \frac{4 c}{3} - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 + \frac{80}{3} + \frac{4 c}{3} - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Multiply $- - \frac{4 c}{3}$ .

Multiply $- 1$ by $- 1$ .

$a = 10 + \frac{80}{3} + 1 (\frac{4 c}{3}) - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

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

$a = 10 + \frac{80}{3} + \frac{4 c}{3} - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 + \frac{80}{3} + \frac{4 c}{3} - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 10 + \frac{80}{3} + \frac{4 c}{3} - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

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

$a = 10 \cdot \frac{3}{3} + \frac{80}{3} + \frac{4 c}{3} - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

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

$a = \frac{10 \cdot 3}{3} + \frac{80}{3} + \frac{4 c}{3} - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Combine the numerators over the common denominator.

$a = \frac{10 \cdot 3 + 80}{3} + \frac{4 c}{3} - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Simplify the numerator.

Multiply $10$ by $3$ .

$a = \frac{30 + 80}{3} + \frac{4 c}{3} - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Add $30$ and $80$ .

$a = \frac{110}{3} + \frac{4 c}{3} - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = \frac{110}{3} + \frac{4 c}{3} - c$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

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

$a = \frac{110}{3} + \frac{4 c}{3} - c \cdot \frac{3}{3}$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

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

$a = \frac{110}{3} + \frac{4 c}{3} + \frac{- c \cdot 3}{3}$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Combine the numerators over the common denominator.

$a = \frac{110}{3} + \frac{4 c - c \cdot 3}{3}$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Combine the numerators over the common denominator.

$a = \frac{110 + 4 c - c \cdot 3}{3}$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Multiply $3$ by $- 1$ .

$a = \frac{110 + 4 c - 3 c}{3}$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Subtract $3 c$ from $4 c$ .

$a = \frac{110 + c}{3}$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = \frac{110 + c}{3}$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = \frac{110 + c}{3}$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = \frac{110 + c}{3}$

$50 = \frac{280 - c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Solve for $c$ in $50 = \frac{280 - c}{3}$ .

Rewrite the equation as $\frac{280 - c}{3} = 50$ .

$\frac{280 - c}{3} = 50$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Multiply both sides by $3$ .

$\frac{280 - c}{3} \cdot 3 = 50 \cdot 3$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Simplify.

Simplify the left side.

Simplify $\frac{280 - c}{3} \cdot 3$ .

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{280 - c}{3} \cdot 3 = 50 \cdot 3$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Rewrite the expression.

$280 - c = 50 \cdot 3$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$280 - c = 50 \cdot 3$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Reorder $280$ and $- c$ .

$- c + 280 = 50 \cdot 3$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$- c + 280 = 50 \cdot 3$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$- c + 280 = 50 \cdot 3$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Simplify the right side.

Multiply $50$ by $3$ .

$- c + 280 = 150$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$- c + 280 = 150$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$- c + 280 = 150$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Solve for $c$ .

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

Subtract $280$ from both sides of the equation.

$- c = 150 - 280$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Subtract $280$ from $150$ .

$- c = - 130$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$- c = - 130$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Divide each term in $- c = - 130$ by $- 1$ and simplify.

Divide each term in $- c = - 130$ by $- 1$ .

$\frac{- c}{- 1} = \frac{- 130}{- 1}$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{c}{1} = \frac{- 130}{- 1}$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Divide $c$ by $1$ .

$c = \frac{- 130}{- 1}$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$c = \frac{- 130}{- 1}$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Simplify the right side.

Divide $- 130$ by $- 1$ .

$c = 130$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$c = 130$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$c = 130$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$c = 130$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$c = 130$

$a = \frac{110 + c}{3}$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Replace all occurrences of $c$ with $130$ in each equation.

Replace all occurrences of $c$ in $a = \frac{110 + c}{3}$ with $130$ .

$a = \frac{110 + 130}{3}$

$c = 130$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Simplify $a = \frac{110 + 130}{3}$ .

Simplify the left side.

Remove parentheses.

$a = \frac{110 + 130}{3}$

$c = 130$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = \frac{110 + 130}{3}$

$c = 130$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Simplify the right side.

Simplify $\frac{110 + 130}{3}$ .

Add $110$ and $130$ .

$a = \frac{240}{3}$

$c = 130$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Divide $240$ by $3$ .

$a = 80$

$c = 130$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 80$

$c = 130$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 80$

$c = 130$

$b = - \frac{80}{3} - \frac{4 c}{3}$

$a = 80$

$c = 130$

$b = - \frac{80}{3} - \frac{4 c}{3}$

Replace all occurrences of $c$ in $b = - \frac{80}{3} - \frac{4 c}{3}$ with $130$ .

$b = - \frac{80}{3} - \frac{4 (130)}{3}$

$a = 80$

$c = 130$

Simplify the right side.

Simplify $- \frac{80}{3} - \frac{4 (130)}{3}$ .

Combine the numerators over the common denominator.

$b = \frac{- 80 - 4 \cdot 130}{3}$

$a = 80$

$c = 130$

Simplify the expression.

Multiply $- 4$ by $130$ .

$b = \frac{- 80 - 520}{3}$

$a = 80$

$c = 130$

Subtract $520$ from $- 80$ .

$b = \frac{- 600}{3}$

$a = 80$

$c = 130$

Divide $- 600$ by $3$ .

$b = - 200$

$a = 80$

$c = 130$

$b = - 200$

$a = 80$

$c = 130$

$b = - 200$

$a = 80$

$c = 130$

$b = - 200$

$a = 80$

$c = 130$

$b = - 200$

$a = 80$

$c = 130$

List all of the solutions.

$b = - 200, a = 80, c = 130$

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

$y = 80 x^{2} - 200 x + 130$

Find the Equation of the Parabola (1,10) , (2,50) , (3,250)

Related Questions

Support LCMGCF.com by disabling your adblocker.