Solve by Completing the Square x^2-12x-5=7

$x^{2} - 12 x - 5 = 7$

Step 1

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

Add $5$ to both sides of the equation.

$x^{2} - 12 x = 7 + 5$

Add $7$ and $5$ .

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

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.

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

Step 4

Simplify the equation.

Simplify the left side.

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

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

Simplify the right side.

Simplify $12 + {(- 6)}^{2}$ .

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

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

Add $12$ and $36$ .

$x^{2} - 12 x + 36 = 48$

Step 5

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

${(x - 6)}^{2} = 48$

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.

$x - 6 = \pm \sqrt{48}$

Simplify $\pm \sqrt{48}$ .

Rewrite $48$ as $4^{2} \cdot 3$ .

Factor $16$ out of $48$ .

$x - 6 = \pm \sqrt{16 (3)}$

Rewrite $16$ as $4^{2}$ .

$x - 6 = \pm \sqrt{4^{2} \cdot 3}$

Pull terms out from under the radical.

$x - 6 = \pm 4 \sqrt{3}$

Add $6$ to both sides of the equation.

$x = \pm 4 \sqrt{3} + 6$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm 4 \sqrt{3} + 6$

Decimal Form:

$x = 12.92820323 \dots, - 0.92820323 \dots$

Solve by Completing the Square x^2-12x-5=7

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