Chapter 10 - Quadratic Equations - 10.4 - Solving Quadratic Equations - Which Method? - Problem Set 10.4 - Page 457: 22

{$-12,5$}

Factoring the equation and setting the factors equal to zero, we obtain: $y^{2}+7y=60$ $y^{2}+7y-60=0$ $y^{2}-5y+12y-60=0$ $y(y-5)+12(y-5)=0$ $(y-5)(y+12)=0$ $(y-5)=0$ or $(y+12)=0$ $y=5$ or $y=-12$ Therefore, the solution set is {$-12,5$}.