Solve Using the Quadratic Formula 4x^2-7x-15=0

$4 x^{2} - 7 x - 15 = 0$

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 $a = 4$ , $b = - 7$ , and $c = - 15$ into the quadratic formula and solve for $x$ .

$\frac{7 \pm \sqrt{{(- 7)}^{2} - 4 \cdot (4 \cdot - 15)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 7$ to the power of $2$ .

$x = \frac{7 \pm \sqrt{49 - 4 \cdot 4 \cdot - 15}}{2 \cdot 4}$

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

Multiply $- 4$ by $4$ .

$x = \frac{7 \pm \sqrt{49 - 16 \cdot - 15}}{2 \cdot 4}$

Multiply $- 16$ by $- 15$ .

$x = \frac{7 \pm \sqrt{49 + 240}}{2 \cdot 4}$

Add $49$ and $240$ .

$x = \frac{7 \pm \sqrt{289}}{2 \cdot 4}$

Rewrite $289$ as $17^{2}$ .

$x = \frac{7 \pm \sqrt{17^{2}}}{2 \cdot 4}$

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

$x = \frac{7 \pm 17}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{7 \pm 17}{8}$

Step 4

The final answer is the combination of both solutions.

$x = 3, - \frac{5}{4}$