Question 144312
 7=24y^2+8y^3

let t=y^2...7=24t+8t^2......did I do this correctly?

No. Close but no cigar

let t=y^2...7=24t+8t^2. if you put y^2 back in to t, you get 7=24y^2+8y^4.

0 = 8y^3 + 24y^2 -7
Try this ? --> http://www.1728.com/cubic2.htm
http://www.1728.com/cubic.htm

Not pretty.