## Intermediate Algebra: Connecting Concepts through Application

The solution set is $\left\{-5-\sqrt{10}, -5+\sqrt{10}\right\}$.
Subtract $21$ from each side to obtain: $$-3(y+5)^2=-30$$ Divide $-3$ to both sides to obtain: $$(y+5)^2=10$$ Take the square root of both sides to obtain: \begin{align*} \sqrt{(y+5)^2}&=\pm\sqrt{10}\\ y+5&=\pm\sqrt{10}\\ y&=-5\pm\sqrt{10} \end{align*} Checking: \begin{align*} -3(-5-\sqrt{10}+5)^2+21&=-9\\ -3(-\sqrt{10}^2+21&=-9\\ -3(10)+21&=-9\\ -30+21&=-9\\ -9&\stackrel{\checkmark}=-9 \end{align*} \begin{align*} -3(-5+\sqrt{10}+5)^2+21&=-9\\ -3(\sqrt{10}^2+21&=-9\\ -3(10)+21&=-9\\ -30+21&=-9\\ -9&\stackrel{\checkmark}=-9 \end{align*} Therefiore, the solution set is $\left\{-5-\sqrt{10}, -5+\sqrt{10}\right\}$.