Question

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

Answer 1

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