先化简,再求值：(y-2)(2y²-3y+9)-y(y²-2y+15),其中Y²－4Y-4

y²-4y-4=0.
y²-4y=4
(y-2)(2y^2-3y+9)-y(y^2-2y+15)
=(y-2)(2y^2-8y+5y+9)-y(y^2-4y+2y+15)
=(y-2)[2(y^2-4y)+5y+9]-y(y^2-4y+2y+15)
=(y-2)[2*4+5y+9]-y(4+2y+15)
=(y-2)(5y+17)-y(2y+19)
=5y^2+7y-34-2y^2-19y
=5y^2-2y^2-19y+7y-34
=3y^2-12y-34
=3(y²-4y)-34
=3*4-34
=12-34
=-22