## Calculus with Applications (10th Edition)

$0$ and $4$
Divide the equation by 12, $12y^{2}-48y=0\qquad.../\div 12$ $y^{2}-4y=0\qquad$ ... factor out y $y(y-4)=0$ By the zero product principle, $y=0$ or $y-4=0$ $y=0$ or $y=4$ The solutions are $0$ and $4$.