Solve by Completing the Square x^2-12x-4=0

$x^{2} - 12 x - 4 = 0$

Step 1

Add $4$ to both sides of the equation.

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

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 $b$ .

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

Step 3

Add the term to each side of the equation.

$x^{2} - 12 x + {(- 6)}^{2} = 4 + {(- 6)}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

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

$x^{2} - 12 x + 36 = 4 + {(- 6)}^{2}$

Simplify the right side.

Simplify $4 + {(- 6)}^{2}$ .

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

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

Add $4$ and $36$ .

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

Step 5

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

${(x - 6)}^{2} = 40$

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.

$x - 6 = \pm \sqrt{40}$

Simplify $\pm \sqrt{40}$ .

Rewrite $40$ as $2^{2} \cdot 10$ .

Factor $4$ out of $40$ .

$x - 6 = \pm \sqrt{4 (10)}$

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

$x - 6 = \pm \sqrt{2^{2} \cdot 10}$

Pull terms out from under the radical.

$x - 6 = \pm 2 \sqrt{10}$

Add $6$ to both sides of the equation.

$x = \pm 2 \sqrt{10} + 6$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm 2 \sqrt{10} + 6$

Decimal Form:

$x = 12.32455532 \dots, - 0.32455532 \dots$