Find the Roots (Zeros) x^2-2x-2=0

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

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 2$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 1 \cdot - 2$ .

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 2$ .

$x = \frac{2 \pm \sqrt{4 + 8}}{2 \cdot 1}$

Add $4$ and $8$ .

$x = \frac{2 \pm \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{2 \pm \sqrt{4 (3)}}{2 \cdot 1}$

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

$x = \frac{2 \pm \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{3}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{3}}{2}$ .

$x = 1 \pm \sqrt{3}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = 1 \pm \sqrt{3}$

Decimal Form:

$x = 2.73205080 \dots, - 0.73205080 \dots$

Step 5