16y^3+48y^2=-36y

I need help setting up the problem.

16y^3+48y^2=-36y

First divide each term by 4 and "move" the term from the right side to the left

4y^3 + 12y^2 + 9y
factor out a y
y(4y^2 + 12y + 9) = 0
I see a perfect square
y(2y + 3)^2 = 0
y = 0 or 2y+3 = 0

y = 0 or y = -3/2