Answer:
[tex]\dfrac{3(y-5)}{2}[/tex]
Step-by-step explanation:
[tex]\dfrac{2y^2-6y-20}{4y+12}\div \dfrac{y^2+5y+6}{3y^2+18y+27}=\\\\\dfrac{2(y-5)(y+2)}{4(y+3)}\cdot \dfrac{3(y+3)^2}{(y+3)(y+2)}=\\\\\dfrac{6(y-5)(y+2)(y+3)(y+3)}{4(y+2)(y+3)(y+3)}=\\\\\dfrac{3(y-5)}{2}[/tex]
Hope this helps!
