Solve Using the Quadratic Formula a^2+4a+8=0

$a^{2} + 4 a + 8 = 0$

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 $a = 1$ , $b = 4$ , and $c = 8$ into the quadratic formula and solve for $a$ .

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

Step 3

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$a = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 8$ .

Multiply $- 4$ by $1$ .

$a = \frac{- 4 \pm \sqrt{16 - 4 \cdot 8}}{2 \cdot 1}$

Multiply $- 4$ by $8$ .

$a = \frac{- 4 \pm \sqrt{16 - 32}}{2 \cdot 1}$

Subtract $32$ from $16$ .

$a = \frac{- 4 \pm \sqrt{- 16}}{2 \cdot 1}$

Rewrite $- 16$ as $- 1 (16)$ .

$a = \frac{- 4 \pm \sqrt{- 1 \cdot 16}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$a = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{16}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$a = \frac{- 4 \pm i \cdot \sqrt{16}}{2 \cdot 1}$

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

$a = \frac{- 4 \pm i \cdot \sqrt{4^{2}}}{2 \cdot 1}$

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

$a = \frac{- 4 \pm i \cdot 4}{2 \cdot 1}$

Move $4$ to the left of $i$ .

$a = \frac{- 4 \pm 4 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$a = \frac{- 4 \pm 4 i}{2}$

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

$a = - 2 \pm 2 i$