Solve Using the Quadratic Formula r^2+8r+15=0

$r^{2} + 8 r + 15 = 0$

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 $a = 1$ , $b = 8$ , and $c = 15$ into the quadratic formula and solve for $r$ .

$\frac{- 8 \pm \sqrt{8^{2} - 4 \cdot (1 \cdot 15)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$r = \frac{- 8 \pm \sqrt{64 - 4 \cdot 1 \cdot 15}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 15$ .

Multiply $- 4$ by $1$ .

$r = \frac{- 8 \pm \sqrt{64 - 4 \cdot 15}}{2 \cdot 1}$

Multiply $- 4$ by $15$ .

$r = \frac{- 8 \pm \sqrt{64 - 60}}{2 \cdot 1}$

Subtract $60$ from $64$ .

$r = \frac{- 8 \pm \sqrt{4}}{2 \cdot 1}$

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

$r = \frac{- 8 \pm \sqrt{2^{2}}}{2 \cdot 1}$

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

$r = \frac{- 8 \pm 2}{2 \cdot 1}$

Multiply $2$ by $1$ .

$r = \frac{- 8 \pm 2}{2}$

Simplify $\frac{- 8 \pm 2}{2}$ .

$r = - 4 \pm 1$

Step 4

The final answer is the combination of both solutions.

$r = - 3, - 5$