Solve Using the Quadratic Formula 4x^2-12x-16=0

$4 x^{2} - 12 x - 16 = 0$

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 $a = 4$ , $b = - 12$ , and $c = - 16$ into the quadratic formula and solve for $x$ .

$\frac{12 \pm \sqrt{{(- 12)}^{2} - 4 \cdot (4 \cdot - 16)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 12$ to the power of $2$ .

$x = \frac{12 \pm \sqrt{144 - 4 \cdot 4 \cdot - 16}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 16$ .

Multiply $- 4$ by $4$ .

$x = \frac{12 \pm \sqrt{144 - 16 \cdot - 16}}{2 \cdot 4}$

Multiply $- 16$ by $- 16$ .

$x = \frac{12 \pm \sqrt{144 + 256}}{2 \cdot 4}$

Add $144$ and $256$ .

$x = \frac{12 \pm \sqrt{400}}{2 \cdot 4}$

Rewrite $400$ as $20^{2}$ .

$x = \frac{12 \pm \sqrt{20^{2}}}{2 \cdot 4}$

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

$x = \frac{12 \pm 20}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{12 \pm 20}{8}$

Simplify $\frac{12 \pm 20}{8}$ .

$x = \frac{3 \pm 5}{2}$

Step 4

The final answer is the combination of both solutions.

$x = 4, - 1$