Solve Using the Quadratic Formula 4x^2+16x+15=0

$4 x^{2} + 16 x + 15 = 0$

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 $a = 4$ , $b = 16$ , and $c = 15$ into the quadratic formula and solve for $x$ .

$\frac{- 16 \pm \sqrt{16^{2} - 4 \cdot (4 \cdot 15)}}{2 \cdot 4}$

Step 3

Simplify.

Simplify the numerator.

Raise $16$ to the power of $2$ .

$x = \frac{- 16 \pm \sqrt{256 - 4 \cdot 4 \cdot 15}}{2 \cdot 4}$

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

Multiply $- 4$ by $4$ .

$x = \frac{- 16 \pm \sqrt{256 - 16 \cdot 15}}{2 \cdot 4}$

Multiply $- 16$ by $15$ .

$x = \frac{- 16 \pm \sqrt{256 - 240}}{2 \cdot 4}$

Subtract $240$ from $256$ .

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

Rewrite $16$ as $4^{2}$ .

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

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

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

Multiply $2$ by $4$ .

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

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

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

Step 4

The final answer is the combination of both solutions.

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