Solve by Completing the Square x^2-12x=-32

$x^{2} - 12 x = - 32$

Step 1

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

${(\frac{b}{2})}^{2} = {(- 6)}^{2}$

Step 2

Add the term to each side of the equation.

$x^{2} - 12 x + {(- 6)}^{2} = - 32 + {(- 6)}^{2}$

Step 3

Simplify the equation.

Simplify the left side.

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

$x^{2} - 12 x + 36 = - 32 + {(- 6)}^{2}$

Simplify the right side.

Simplify $- 32 + {(- 6)}^{2}$ .

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

$x^{2} - 12 x + 36 = - 32 + 36$

Add $- 32$ and $36$ .

$x^{2} - 12 x + 36 = 4$

Step 4

Factor the perfect trinomial square into ${(x - 6)}^{2}$ .

${(x - 6)}^{2} = 4$

Step 5

Solve the equation for $x$ .

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

$x - 6 = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

Rewrite $4$ as $2^{2}$ .

$x - 6 = \pm \sqrt{2^{2}}$

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

$x - 6 = \pm 2$

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.

$x - 6 = 2$

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

Add $6$ to both sides of the equation.

$x = 2 + 6$

Add $2$ and $6$ .

$x = 8$

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

$x - 6 = - 2$

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

Add $6$ to both sides of the equation.

$x = - 2 + 6$

Add $- 2$ and $6$ .

$x = 4$

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

$x = 8, 4$

Solve by Completing the Square x^2-12x=-32

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