Question

Solve the given equation.

${\left(2y+3\right)}^2+3\left(2y+3\right)-28=0$

Short answer

The solutions are$y=-5$and$y=\frac{1}{2}$.

Step-by-step solution

Step 1. Given information.

The given equation is${\left(2y+3\right)}^2+3\left(2y+3\right)-28=0$.

Step 2. Simplify the equation.

$\begin{array}{rcl}{\left(2y+3\right)}^2+3\left(2y+3\right)-28& =& 0\\ {\left(2y+3\right)}^2+7\left(2y+3\right)-4\left(2y+3\right)-28& =& 0\\ \left(2y+3\right)\left(2y+3+7\right)-4\left(2y+3+7\right)& =& 0\\ \left(2y+3+7\right)\left(2y+3-4\right)& =& 0\end{array}$

Step 3. Solve the equation for y.

$\begin{array}{rcl}2y+3+7& =& 0\\ 2y+10& =& 0\\ 2y& =& -10\\ y& =& -\frac{10}{2}\\ y& =& -5\end{array}$ and $\begin{array}{rcl}2y+3-4& =& 0\\ 2y-1& =& 0\\ 2y& =& 1\\ y& =& \frac{1}{2}\end{array}$

Step 4. Simplified answer.

Hence, the required solutions are$y=-5$and$y=\frac{1}{2}$.