Solve Using the Quadratic Formula r^2+7r+12=0

$r^{2} + 7 r + 12 = 0$

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 $a = 1$ , $b = 7$ , and $c = 12$ into the quadratic formula and solve for $r$ .

$\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (1 \cdot 12)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$r = \frac{- 7 \pm \sqrt{49 - 4 \cdot 1 \cdot 12}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 12$ .

Multiply $- 4$ by $1$ .

$r = \frac{- 7 \pm \sqrt{49 - 4 \cdot 12}}{2 \cdot 1}$

Multiply $- 4$ by $12$ .

$r = \frac{- 7 \pm \sqrt{49 - 48}}{2 \cdot 1}$

Subtract $48$ from $49$ .

$r = \frac{- 7 \pm \sqrt{1}}{2 \cdot 1}$

Any root of $1$ is $1$ .

$r = \frac{- 7 \pm 1}{2 \cdot 1}$

Multiply $2$ by $1$ .

$r = \frac{- 7 \pm 1}{2}$

Step 4

The final answer is the combination of both solutions.

$r = - 3, - 4$