Solve Using the Quadratic Formula 2x^2+2x-2=0

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

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

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $2$ .

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

Multiply $- 8$ by $- 2$ .

$x = \frac{- 2 \pm \sqrt{4 + 16}}{2 \cdot 2}$

Add $4$ and $16$ .

$x = \frac{- 2 \pm \sqrt{20}}{2 \cdot 2}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$x = \frac{- 2 \pm \sqrt{4 (5)}}{2 \cdot 2}$

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

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 2}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{5}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 2 \pm 2 \sqrt{5}}{4}$

Simplify $\frac{- 2 \pm 2 \sqrt{5}}{4}$ .

$x = \frac{- 1 \pm \sqrt{5}}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 1 \pm \sqrt{5}}{2}$

Decimal Form:

$x = 0.61803398 \dots, - 1.61803398 \dots$