Solve Using the Quadratic Formula -2x=x^2-6

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

Move all the expressions to the left side of the equation.

Subtract $x^{2}$ from both sides of the equation.

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

Add $6$ to both sides of the equation.

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

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 $c = 6$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (- 1 \cdot 6)}}{2 \cdot - 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 1 \cdot 6}}{2 \cdot - 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{2 \pm \sqrt{4 + 4 \cdot 6}}{2 \cdot - 1}$

Multiply $4$ by $6$ .

$x = \frac{2 \pm \sqrt{4 + 24}}{2 \cdot - 1}$

Add $4$ and $24$ .

$x = \frac{2 \pm \sqrt{28}}{2 \cdot - 1}$

Rewrite $28$ as $2^{2} \cdot 7$ .

Factor $4$ out of $28$ .

$x = \frac{2 \pm \sqrt{4 (7)}}{2 \cdot - 1}$

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

$x = \frac{2 \pm \sqrt{2^{2} \cdot 7}}{2 \cdot - 1}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{7}}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{2 \pm 2 \sqrt{7}}{- 2}$

Simplify $\frac{2 \pm 2 \sqrt{7}}{- 2}$ .

$x = \frac{1 \pm \sqrt{7}}{- 1}$

Move the negative one from the denominator of $\frac{1 \pm \sqrt{7}}{- 1}$ .

$x = - 1 \cdot (1 \pm \sqrt{7})$

Rewrite $- 1 \cdot (1 \pm \sqrt{7})$ as $- (1 \pm \sqrt{7})$ .

$x = - (1 \pm \sqrt{7})$

The result can be shown in multiple forms.

Exact Form:

$x = - (1 \pm \sqrt{7})$

Decimal Form:

$x = - 3.64575131 \dots, 1.64575131 \dots$