# How do you solve b^2=-3b?

Mar 9, 2017

See the entire solution process below:

#### Explanation:

First, add $\textcolor{red}{3 b}$ to each side of the equation to put the equation in standard form:

${b}^{2} + \textcolor{red}{3 b} = - 3 b + \textcolor{red}{3 b}$

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

Next, factor a $b$ out of each term on the left:

$\left(b \cdot b\right) + \left(3 \cdot b\right) = 0$

$b \left(b + 3\right) = 0$

Now, solve each term on the left for $0$ to determine all of the solutions:

Solution 1:

$b = 0$

Solution 2:

$b + 3 = 0$

$b + 3 - \textcolor{red}{3} = 0 - \textcolor{red}{3}$

$b + 0 = - 3$

$b = - 3$

The Solutions Are:

$b = \left[- 3 , 0\right\}$