Solve Using the Quadratic Formula 2-x^2=-11

$2 - x^{2} = - 11$

Move all terms to the left side of the equation and simplify.

Add $11$ to both sides of the equation.

$2 - x^{2} + 11 = 0$

Add $2$ and $11$ .

$- x^{2} + 13 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = - 1$ , $b = 0$ , and $c = 13$ into the quadratic formula and solve for $x$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (- 1 \cdot 13)}}{2 \cdot - 1}$

Simplify.

Simplify the numerator.

Raising $0$ to any positive power yields $0$ .

$x = \frac{0 \pm \sqrt{0 - 4 \cdot - 1 \cdot 13}}{2 \cdot - 1}$

Multiply $- 4$ by $- 1$ .

$x = \frac{0 \pm \sqrt{0 + 4 \cdot 13}}{2 \cdot - 1}$

Multiply $4$ by $13$ .

$x = \frac{0 \pm \sqrt{0 + 52}}{2 \cdot - 1}$

Add $0$ and $52$ .

$x = \frac{0 \pm \sqrt{52}}{2 \cdot - 1}$

Rewrite $52$ as $2^{2} \cdot 13$ .

Factor $4$ out of $52$ .

$x = \frac{0 \pm \sqrt{4 (13)}}{2 \cdot - 1}$

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

$x = \frac{0 \pm \sqrt{2^{2} \cdot 13}}{2 \cdot - 1}$

Pull terms out from under the radical.

$x = \frac{0 \pm 2 \sqrt{13}}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{0 \pm 2 \sqrt{13}}{- 2}$

Simplify $\frac{0 \pm 2 \sqrt{13}}{- 2}$ .

$x = \pm \sqrt{13}$

The result can be shown in multiple forms.

Exact Form:

$x = \pm \sqrt{13}$

Decimal Form:

$x = 3.60555127 \dots, - 3.60555127 \dots$