Solve Using the Quadratic Formula r^2+9r+20=0

$r^{2} + 9 r + 20 = 0$

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 $a = 1$ , $b = 9$ , and $c = 20$ into the quadratic formula and solve for $r$ .

$\frac{- 9 \pm \sqrt{9^{2} - 4 \cdot (1 \cdot 20)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $9$ to the power of $2$ .

$r = \frac{- 9 \pm \sqrt{81 - 4 \cdot 1 \cdot 20}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 20$ .

Multiply $- 4$ by $1$ .

$r = \frac{- 9 \pm \sqrt{81 - 4 \cdot 20}}{2 \cdot 1}$

Multiply $- 4$ by $20$ .

$r = \frac{- 9 \pm \sqrt{81 - 80}}{2 \cdot 1}$

Subtract $80$ from $81$ .

$r = \frac{- 9 \pm \sqrt{1}}{2 \cdot 1}$

Any root of $1$ is $1$ .

$r = \frac{- 9 \pm 1}{2 \cdot 1}$

Multiply $2$ by $1$ .

$r = \frac{- 9 \pm 1}{2}$

Step 4

The final answer is the combination of both solutions.

$r = - 4, - 5$