63x=3x^2+30x^3

I'm assuming you are trying to solve for x?

Okay, your first step is to get all of the variables on one side. So subtract 63x from both sides and you get
30x^3+3x^2-63x = 0

The next thing would be to look at what you can "pull out" of all three "sets" of numbers. Both three and x go into everything on the left side of the equation, so you can factor that out and you get
3x(10x^2+x-21) = 0.

Your next step is to use FOIL to separate the part in the parentheses. That stands for Firsts, Outers, Inners, Lasts. Once you play around with the numbers a bit you see that when you multiply (5x-7)(2x+3), you get 10x^2+x-21.

Your final equation would be
3x(5x-7)(2x+3)=0. So, your solutions would be x=0, x=5/7, and x=-2/3

Hope that helps!