Solve Using the Quadratic Formula 4x^2-8x-12=0

$4 x^{2} - 8 x - 12 = 0$

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 $a = 4$ , $b = - 8$ , and $c = - 12$ into the quadratic formula and solve for $x$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (4 \cdot - 12)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 8$ to the power of $2$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 4 \cdot - 12}}{2 \cdot 4}$

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

Multiply $- 4$ by $4$ .

$x = \frac{8 \pm \sqrt{64 - 16 \cdot - 12}}{2 \cdot 4}$

Multiply $- 16$ by $- 12$ .

$x = \frac{8 \pm \sqrt{64 + 192}}{2 \cdot 4}$

Add $64$ and $192$ .

$x = \frac{8 \pm \sqrt{256}}{2 \cdot 4}$

Rewrite $256$ as $16^{2}$ .

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

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

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

Multiply $2$ by $4$ .

$x = \frac{8 \pm 16}{8}$

Simplify $\frac{8 \pm 16}{8}$ .

$x = 1 \pm 2$

Step 4

The final answer is the combination of both solutions.

$x = 3, - 1$