Solve for x 2x^2-2x-1=0

$2 x^{2} - 2 x - 1 = 0$

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 $c = - 1$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $2$ .

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

Multiply $- 8$ by $- 1$ .

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

Add $4$ and $8$ .

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

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

Factor $4$ out of $12$ .

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

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

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

Pull terms out from under the radical.

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

Multiply $2$ by $2$ .

$x = \frac{2 \pm 2 \sqrt{3}}{4}$

Simplify $\frac{2 \pm 2 \sqrt{3}}{4}$ .

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

Step 4

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{3}}{2}, \frac{1 - \sqrt{3}}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1 + \sqrt{3}}{2}, \frac{1 - \sqrt{3}}{2}$

Decimal Form:

$x = 1.36602540 \dots, - 0.36602540 \dots$