Solve for x -2x^2-4x=6

$- 2 x^{2} - 4 x = 6$

Subtract $6$ from both sides of the equation.

$- 2 x^{2} - 4 x - 6 = 0$

Factor $- 2$ out of $- 2 x^{2} - 4 x - 6$ .

Factor $- 2$ out of $- 4 x$ .

$- 2 x^{2} - 2 (2 x) - 6 = 0$

Factor $- 2$ out of $- 6$ .

$- 2 x^{2} - 2 (2 x) - 2 \cdot 3 = 0$

Factor $- 2$ out of $- 2 (x^{2}) - 2 (2 x)$ .

$- 2 (x^{2} + 2 x) - 2 \cdot 3 = 0$

Factor $- 2$ out of $- 2 (x^{2} + 2 x) - 2 (3)$ .

$- 2 (x^{2} + 2 x + 3) = 0$

Divide each term in $- 2 (x^{2} + 2 x + 3) = 0$ by $- 2$ and simplify.

Divide each term in $- 2 (x^{2} + 2 x + 3) = 0$ by $- 2$ .

$\frac{- 2 (x^{2} + 2 x + 3)}{- 2} = \frac{0}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 (x^{2} + 2 x + 3)}{- 2} = \frac{0}{- 2}$

Divide $x^{2} + 2 x + 3$ by $1$ .

$x^{2} + 2 x + 3 = \frac{0}{- 2}$

Simplify the right side.

Divide $0$ by $- 2$ .

$x^{2} + 2 x + 3 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 2$ , and $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot 3)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $3$ .

$x = \frac{- 2 \pm \sqrt{4 - 12}}{2 \cdot 1}$

Subtract $12$ from $4$ .

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

Rewrite $- 8$ as $- 1 (8)$ .

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

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

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

Rewrite $\sqrt{- 1}$ as $i$ .

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

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

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

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

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

Pull terms out from under the radical.

$x = \frac{- 2 \pm i \cdot (2 \sqrt{2})}{2 \cdot 1}$

Move $2$ to the left of $i$ .

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

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 i \sqrt{2}}{2}$

Simplify $\frac{- 2 \pm 2 i \sqrt{2}}{2}$ .

$x = - 1 \pm i \sqrt{2}$

The final answer is the combination of both solutions.

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

Solve for x -2x^2-4x=6

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