Solve by Completing the Square 12x^2+7x=10

$12 x^{2} + 7 x = 10$

Step 1

Divide each term in $12 x^{2} + 7 x = 10$ by $12$ and simplify.

Divide each term in $12 x^{2} + 7 x = 10$ by $12$ .

$\frac{12 x^{2}}{12} + \frac{7 x}{12} = \frac{10}{12}$

Simplify the left side.

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12 x^{2}}{12} + \frac{7 x}{12} = \frac{10}{12}$

Divide $x^{2}$ by $1$ .

$x^{2} + \frac{7 x}{12} = \frac{10}{12}$

Simplify the right side.

Cancel the common factor of $10$ and $12$ .

Factor $2$ out of $10$ .

$x^{2} + \frac{7 x}{12} = \frac{2 (5)}{12}$

Cancel the common factors.

Factor $2$ out of $12$ .

$x^{2} + \frac{7 x}{12} = \frac{2 \cdot 5}{2 \cdot 6}$

Cancel the common factor.

$x^{2} + \frac{7 x}{12} = \frac{2 \cdot 5}{2 \cdot 6}$

Rewrite the expression.

$x^{2} + \frac{7 x}{12} = \frac{5}{6}$

Step 2

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(\frac{7}{24})}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} + \frac{7 x}{12} + {(\frac{7}{24})}^{2} = \frac{5}{6} + {(\frac{7}{24})}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{7}{24}$ .

$x^{2} + \frac{7 x}{12} + \frac{7^{2}}{24^{2}} = \frac{5}{6} + {(\frac{7}{24})}^{2}$

Raise $7$ to the power of $2$ .

$x^{2} + \frac{7 x}{12} + \frac{49}{24^{2}} = \frac{5}{6} + {(\frac{7}{24})}^{2}$

Raise $24$ to the power of $2$ .

$x^{2} + \frac{7 x}{12} + \frac{49}{576} = \frac{5}{6} + {(\frac{7}{24})}^{2}$

Simplify the right side.

Simplify $\frac{5}{6} + {(\frac{7}{24})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{7}{24}$ .

$x^{2} + \frac{7 x}{12} + \frac{49}{576} = \frac{5}{6} + \frac{7^{2}}{24^{2}}$

Raise $7$ to the power of $2$ .

$x^{2} + \frac{7 x}{12} + \frac{49}{576} = \frac{5}{6} + \frac{49}{24^{2}}$

Raise $24$ to the power of $2$ .

$x^{2} + \frac{7 x}{12} + \frac{49}{576} = \frac{5}{6} + \frac{49}{576}$

To write $\frac{5}{6}$ as a fraction with a common denominator, multiply by $\frac{96}{96}$ .

$x^{2} + \frac{7 x}{12} + \frac{49}{576} = \frac{5}{6} \cdot \frac{96}{96} + \frac{49}{576}$

Write each expression with a common denominator of $576$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{5}{6}$ by $\frac{96}{96}$ .

$x^{2} + \frac{7 x}{12} + \frac{49}{576} = \frac{5 \cdot 96}{6 \cdot 96} + \frac{49}{576}$

Multiply $6$ by $96$ .

$x^{2} + \frac{7 x}{12} + \frac{49}{576} = \frac{5 \cdot 96}{576} + \frac{49}{576}$

Combine the numerators over the common denominator.

$x^{2} + \frac{7 x}{12} + \frac{49}{576} = \frac{5 \cdot 96 + 49}{576}$

Simplify the numerator.

Multiply $5$ by $96$ .

$x^{2} + \frac{7 x}{12} + \frac{49}{576} = \frac{480 + 49}{576}$

Add $480$ and $49$ .

$x^{2} + \frac{7 x}{12} + \frac{49}{576} = \frac{529}{576}$

Step 5

Factor the perfect trinomial square into ${(x + \frac{7}{24})}^{2}$ .

${(x + \frac{7}{24})}^{2} = \frac{529}{576}$

Step 6

Solve the equation for $x$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x + \frac{7}{24} = \pm \sqrt{\frac{529}{576}}$

Simplify $\pm \sqrt{\frac{529}{576}}$ .

Rewrite $\sqrt{\frac{529}{576}}$ as $\frac{\sqrt{529}}{\sqrt{576}}$ .

$x + \frac{7}{24} = \pm \frac{\sqrt{529}}{\sqrt{576}}$

Simplify the numerator.

Rewrite $529$ as $23^{2}$ .

$x + \frac{7}{24} = \pm \frac{\sqrt{23^{2}}}{\sqrt{576}}$

Pull terms out from under the radical, assuming positive real numbers.

$x + \frac{7}{24} = \pm \frac{23}{\sqrt{576}}$

Simplify the denominator.

Rewrite $576$ as $24^{2}$ .

$x + \frac{7}{24} = \pm \frac{23}{\sqrt{24^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

$x + \frac{7}{24} = \pm \frac{23}{24}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x + \frac{7}{24} = \frac{23}{24}$

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

Subtract $\frac{7}{24}$ from both sides of the equation.

$x = \frac{23}{24} - \frac{7}{24}$

Combine the numerators over the common denominator.

$x = \frac{23 - 7}{24}$

Subtract $7$ from $23$ .

$x = \frac{16}{24}$

Cancel the common factor of $16$ and $24$ .

Factor $8$ out of $16$ .

$x = \frac{8 (2)}{24}$

Cancel the common factors.

Factor $8$ out of $24$ .

$x = \frac{8 \cdot 2}{8 \cdot 3}$

Cancel the common factor.

$x = \frac{8 \cdot 2}{8 \cdot 3}$

Rewrite the expression.

$x = \frac{2}{3}$

Next, use the negative value of the $\pm$ to find the second solution.

$x + \frac{7}{24} = - \frac{23}{24}$

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

Subtract $\frac{7}{24}$ from both sides of the equation.

$x = - \frac{23}{24} - \frac{7}{24}$

Combine the numerators over the common denominator.

$x = \frac{- 23 - 7}{24}$

Subtract $7$ from $- 23$ .

$x = \frac{- 30}{24}$

Cancel the common factor of $- 30$ and $24$ .

Factor $6$ out of $- 30$ .

$x = \frac{6 (- 5)}{24}$

Cancel the common factors.

Factor $6$ out of $24$ .

$x = \frac{6 \cdot - 5}{6 \cdot 4}$

Cancel the common factor.

$x = \frac{6 \cdot - 5}{6 \cdot 4}$

Rewrite the expression.

$x = \frac{- 5}{4}$

Move the negative in front of the fraction.

$x = - \frac{5}{4}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \frac{2}{3}, - \frac{5}{4}$

Solve by Completing the Square 12x^2+7x=10

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