## Intermediate Algebra: Connecting Concepts through Application

The solution set is $\left\{-3, 0\right\}$.
Factor out $6b$ to obtain: $$6b(b+3)=0$$ Use the Zero-Product Property by equating each factor to zero, then solving each equation to obtain: \begin{align*} 6b&=0 &\text{or}& &b+3=0\\ b&=0 &\text{or}& &b=-3\\ \end{align*} Checking: \begin{align*} 6(0^2)+18(0)&=0\\ 6(0)+0&=0\\ 0&\stackrel{\checkmark}=0 \end{align*} \begin{align*} 6(-3)^2+18(-3)&=0\\ 6(9)+(-54)&=0\\ 0&\stackrel{\checkmark}=0 \end{align*} Therefore, the solution set is $\left\{-3, 0\right\}$.