Solve Using the Quadratic Formula b^2+2b=-20

$b^{2} + 2 b = - 20$

Step 1

Add $20$ to both sides of the equation.

$b^{2} + 2 b + 20 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 1$ , $b = 2$ , and $c = 20$ into the quadratic formula and solve for $b$ .

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

Step 4

Simplify.

Simplify the numerator.

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

$b = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot 20}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$b = \frac{- 2 \pm \sqrt{4 - 4 \cdot 20}}{2 \cdot 1}$

Multiply $- 4$ by $20$ .

$b = \frac{- 2 \pm \sqrt{4 - 80}}{2 \cdot 1}$

Subtract $80$ from $4$ .

$b = \frac{- 2 \pm \sqrt{- 76}}{2 \cdot 1}$

Rewrite $- 76$ as $- 1 (76)$ .

$b = \frac{- 2 \pm \sqrt{- 1 \cdot 76}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (76)}$ as $\sqrt{- 1} \cdot \sqrt{76}$ .

$b = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{76}}{2 \cdot 1}$

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

$b = \frac{- 2 \pm i \cdot \sqrt{76}}{2 \cdot 1}$

Rewrite $76$ as $2^{2} \cdot 19$ .

Factor $4$ out of $76$ .

$b = \frac{- 2 \pm i \cdot \sqrt{4 (19)}}{2 \cdot 1}$

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

$b = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 19}}{2 \cdot 1}$

Pull terms out from under the radical.

$b = \frac{- 2 \pm i \cdot (2 \sqrt{19})}{2 \cdot 1}$

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

$b = \frac{- 2 \pm 2 i \sqrt{19}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$b = \frac{- 2 \pm 2 i \sqrt{19}}{2}$

Simplify $\frac{- 2 \pm 2 i \sqrt{19}}{2}$ .

$b = - 1 \pm i \sqrt{19}$