Solve Using the Quadratic Formula 2x^2-2x-4=0

$2 x^{2} - 2 x - 4 = 0$

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 $c = - 4$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $2$ .

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

Multiply $- 8$ by $- 4$ .

$x = \frac{2 \pm \sqrt{4 + 32}}{2 \cdot 2}$

Add $4$ and $32$ .

$x = \frac{2 \pm \sqrt{36}}{2 \cdot 2}$

Rewrite $36$ as $6^{2}$ .

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

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{2 \pm 6}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{2 \pm 6}{4}$

Simplify $\frac{2 \pm 6}{4}$ .

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

Step 4

The final answer is the combination of both solutions.

$x = 2, - 1$