Solve Using the Quadratic Formula 4x^2-4x-15=0

$4 x^{2} - 4 x - 15 = 0$

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 $a = 4$ , $b = - 4$ , and $c = - 15$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (4 \cdot - 15)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 4 \cdot - 15}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot - 15$ .

Multiply $- 4$ by $4$ .

$x = \frac{4 \pm \sqrt{16 - 16 \cdot - 15}}{2 \cdot 4}$

Multiply $- 16$ by $- 15$ .

$x = \frac{4 \pm \sqrt{16 + 240}}{2 \cdot 4}$

Add $16$ and $240$ .

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

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

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

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

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

Multiply $2$ by $4$ .

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

Simplify $\frac{4 \pm 16}{8}$ .

$x = \frac{1 \pm 4}{2}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{5}{2}, - \frac{3}{2}$