Solve Using the Quadratic Formula 2x^2+2x-12=0

$2 x^{2} + 2 x - 12 = 0$

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 $c = - 12$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (2 \cdot - 12)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 2 \cdot - 12$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 8 \cdot - 12}}{2 \cdot 2}$

Multiply $- 8$ by $- 12$ .

$x = \frac{- 2 \pm \sqrt{4 + 96}}{2 \cdot 2}$

Add $4$ and $96$ .

$x = \frac{- 2 \pm \sqrt{100}}{2 \cdot 2}$

Rewrite $100$ as $10^{2}$ .

$x = \frac{- 2 \pm \sqrt{10^{2}}}{2 \cdot 2}$

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

$x = \frac{- 2 \pm 10}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 2 \pm 10}{4}$

Simplify $\frac{- 2 \pm 10}{4}$ .

$x = \frac{- 1 \pm 5}{2}$

Step 4

The final answer is the combination of both solutions.

$x = 2, - 3$