Solve for y:
(y-(10/y))^2+6(y-(10/y))-27=0
(y-(10/y))^2+6(y-(10/y))-27=0
his is just like the other one: Let u=y-1/y and you have
u^2 + 6u -27 = 0
(u+9)(u-3) = 0
(y - 10/y + 9)(y - 10/y - 3) = 0
(y^2 + 9y - 10)(y^2 - 3y - 10) = 0
(y+10)(y-1)(y-5)(y+3) = 0