Find the Roots (Zeros) x^2-8x-8=-2x

$x^{2} - 8 x - 8 = - 2 x$

Move all terms containing $x$ to the left side of the equation.

Add $2 x$ to both sides of the equation.

$x^{2} - 8 x - 8 + 2 x = 0$

Add $- 8 x$ and $2 x$ .

$x^{2} - 6 x - 8 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 6$ , and $c = - 8$ into the quadratic formula and solve for $x$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot - 8)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot - 8}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{6 \pm \sqrt{36 - 4 \cdot - 8}}{2 \cdot 1}$

Multiply $- 4$ by $- 8$ .

$x = \frac{6 \pm \sqrt{36 + 32}}{2 \cdot 1}$

Add $36$ and $32$ .

$x = \frac{6 \pm \sqrt{68}}{2 \cdot 1}$

Rewrite $68$ as $2^{2} \cdot 17$ .

Factor $4$ out of $68$ .

$x = \frac{6 \pm \sqrt{4 (17)}}{2 \cdot 1}$

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

$x = \frac{6 \pm \sqrt{2^{2} \cdot 17}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{6 \pm 2 \sqrt{17}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{6 \pm 2 \sqrt{17}}{2}$

Simplify $\frac{6 \pm 2 \sqrt{17}}{2}$ .

$x = 3 \pm \sqrt{17}$

The result can be shown in multiple forms.

Exact Form:

$x = 3 \pm \sqrt{17}$

Decimal Form:

$x = 7.12310562 \dots, - 1.12310562 \dots$