Solve for b b^2+3b=-3

$b^{2} + 3 b = - 3$

Step 1

Add $3$ to both sides of the equation.

$b^{2} + 3 b + 3 = 0$

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 $a = 1$ , $b = 3$ , and $c = 3$ into the quadratic formula and solve for $b$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot 3)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $3$ to the power of $2$ .

$b = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 3$ .

Multiply $- 4$ by $1$ .

$b = \frac{- 3 \pm \sqrt{9 - 4 \cdot 3}}{2 \cdot 1}$

Multiply $- 4$ by $3$ .

$b = \frac{- 3 \pm \sqrt{9 - 12}}{2 \cdot 1}$

Subtract $12$ from $9$ .

$b = \frac{- 3 \pm \sqrt{- 3}}{2 \cdot 1}$

Rewrite $- 3$ as $- 1 (3)$ .

$b = \frac{- 3 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$b = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

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

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

Multiply $2$ by $1$ .

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

Step 5

The final answer is the combination of both solutions.

$b = - \frac{3 - i \sqrt{3}}{2}, - \frac{3 + i \sqrt{3}}{2}$