Solve by Completing the Square x^2+12x=40

$x^{2} + 12 x = 40$

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.

$x^{2} + 12 x + {(6)}^{2} = 40 + {(6)}^{2}$

Step 3

Simplify the equation.

Simplify the left side.

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

$x^{2} + 12 x + 36 = 40 + {(6)}^{2}$

Simplify the right side.

Simplify $40 + {(6)}^{2}$ .

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

$x^{2} + 12 x + 36 = 40 + 36$

Add $40$ and $36$ .

$x^{2} + 12 x + 36 = 76$

Step 4

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

${(x + 6)}^{2} = 76$

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.

$x + 6 = \pm \sqrt{76}$

Simplify $\pm \sqrt{76}$ .

Rewrite $76$ as $2^{2} \cdot 19$ .

Factor $4$ out of $76$ .

$x + 6 = \pm \sqrt{4 (19)}$

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

$x + 6 = \pm \sqrt{2^{2} \cdot 19}$

Pull terms out from under the radical.

$x + 6 = \pm 2 \sqrt{19}$

Subtract $6$ from both sides of the equation.

$x = \pm 2 \sqrt{19} - 6$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \pm 2 \sqrt{19} - 6$

Decimal Form:

$x = 2.71779788 \dots, - 14.71779788 \dots$