Question

18(y^2+6y+9)/(y-3)(y+3)12y

Answer 1

$displaystyle frac{18(y^{2}+6x+9) }{12y(y-3)(y+3)}= frac{3(y+3)^{2}}{2y(y-3)(y+3)}= frac{3(y+3)}{2y(y-3)}; \ \ \ y^{2}+6x+9=0 \ D=b^{2}-4ac=36-36=0 \ y_{1}=y_{2}=- frac{b}{2a}=-3$