Solve Using the Quadratic Formula x^2+8x+19=0

$x^{2} + 8 x + 19 = 0$

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 $c = 19$ into the quadratic formula and solve for $x$ .

$\frac{- 8 \pm \sqrt{8^{2} - 4 \cdot (1 \cdot 19)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot 1 \cdot 19}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 19$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 8 \pm \sqrt{64 - 4 \cdot 19}}{2 \cdot 1}$

Multiply $- 4$ by $19$ .

$x = \frac{- 8 \pm \sqrt{64 - 76}}{2 \cdot 1}$

Subtract $76$ from $64$ .

$x = \frac{- 8 \pm \sqrt{- 12}}{2 \cdot 1}$

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{- 8 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{- 8 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

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

$x = \frac{- 8 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{- 8 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

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

$x = \frac{- 8 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 8 \pm i \cdot (2 \sqrt{3})}{2 \cdot 1}$

Move $2$ to the left of $i$ .

$x = \frac{- 8 \pm 2 i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 8 \pm 2 i \sqrt{3}}{2}$

Simplify $\frac{- 8 \pm 2 i \sqrt{3}}{2}$ .

$x = - 4 \pm i \sqrt{3}$