## Intermediate Algebra: Connecting Concepts through Application

$\color{blue}{\left\{-\frac{3}{2}, 0\right\}}$
Factor out $\dfrac{1}{5}x$ to obtain: \begin{align*} \frac{1}{5}x\left(2x+3\right)&=0 \end{align*} Use the Zero-Product Property by equating each factor to zero. Then, solve each equation to obtain: \begin{align*} \frac{1}{5}x&=0 &\text{or}& &2x+3&=0\\\\ x&=0 &\text{or}& &2x=-3\\\\ x&=0 &\text{or}& &x=-\frac{3}{2}\\\\ \end{align*} Therefore, the solution set is $\color{blue}{\left\{-\frac{3}{2}, 0\right\}}$.