Answer
$\frac{(2y+9)(y-2)}{4}$
Work Step by Step
$\frac{2y+9}{4y+12}\times(y^2+y-6)$
$=\frac{2y+9}{4(y+3)}\times(y^2+3y-2y-6)$
$=\frac{2y+9}{4(y+3)}\times[y(y+3)-2(y+3)$
$=\frac{2y+9}{4(y+3)}\times(y-2)(y+3)$
$=\frac{(2y+9)(y-2)(y+3)}{4(y+3)}$
$=\frac{(2y+9)(y-2)}{4}$