Find All Complex Solutions 2x^2-4x=-2

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

Add $2$ to both sides of the equation.

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

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = - 4$ , and $c = 2$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

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

Multiply $- 4$ by $2$ .

$x = \frac{4 \pm \sqrt{16 - 8 \cdot 2}}{2 \cdot 2}$

Multiply $- 8$ by $2$ .

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

Subtract $16$ from $16$ .

$x = \frac{4 \pm \sqrt{0}}{2 \cdot 2}$

Rewrite $0$ as $0^{2}$ .

$x = \frac{4 \pm \sqrt{0^{2}}}{2 \cdot 2}$

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

$x = \frac{4 \pm 0}{2 \cdot 2}$

$4$ plus or minus $0$ is $4$ .

$x = \frac{4}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{4}{4}$

Divide $4$ by $4$ .

$x = 1$

The final answer is the combination of both solutions.

$x = 1$ Double roots