Solve Using the Quadratic Formula k^2+8k+12=0

$k^{2} + 8 k + 12 = 0$

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 $a = 1$ , $b = 8$ , and $c = 12$ into the quadratic formula and solve for $k$ .

$\frac{- 8 \pm \sqrt{8^{2} - 4 \cdot (1 \cdot 12)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$k = \frac{- 8 \pm \sqrt{64 - 4 \cdot 1 \cdot 12}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$k = \frac{- 8 \pm \sqrt{64 - 4 \cdot 12}}{2 \cdot 1}$

Multiply $- 4$ by $12$ .

$k = \frac{- 8 \pm \sqrt{64 - 48}}{2 \cdot 1}$

Subtract $48$ from $64$ .

$k = \frac{- 8 \pm \sqrt{16}}{2 \cdot 1}$

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

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

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

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

Multiply $2$ by $1$ .

$k = \frac{- 8 \pm 4}{2}$

Simplify $\frac{- 8 \pm 4}{2}$ .

$k = - 4 \pm 2$

Step 4

The final answer is the combination of both solutions.

$k = - 2, - 6$