Solve by Completing the Square x^2+12x=16

$x^{2} + 12 x = 16$

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.

$x^{2} + 12 x + {(6)}^{2} = 16 + {(6)}^{2}$

Step 3

Simplify the equation.

Simplify the left side.

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

$x^{2} + 12 x + 36 = 16 + {(6)}^{2}$

Simplify the right side.

Simplify $16 + {(6)}^{2}$ .

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

$x^{2} + 12 x + 36 = 16 + 36$

Add $16$ and $36$ .

$x^{2} + 12 x + 36 = 52$

Step 4

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

${(x + 6)}^{2} = 52$

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.

$x + 6 = \pm \sqrt{52}$

Simplify $\pm \sqrt{52}$ .

Rewrite $52$ as $2^{2} \cdot 13$ .

Factor $4$ out of $52$ .

$x + 6 = \pm \sqrt{4 (13)}$

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

$x + 6 = \pm \sqrt{2^{2} \cdot 13}$

Pull terms out from under the radical.

$x + 6 = \pm 2 \sqrt{13}$

Subtract $6$ from both sides of the equation.

$x = \pm 2 \sqrt{13} - 6$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \pm 2 \sqrt{13} - 6$

Decimal Form:

$x = 1.21110255 \dots, - 13.21110255 \dots$