Solving Separable Equations: How to Solve for Y in Terms of X

[davegillmour]

Ok, I'm having trouble with the last part of this problem. After solving for the constant C, I get:
y^2 - 2y = x^3 + 2x^2 +2x + 3

My question is, how do I solve this in terms of Y? The only instruction my book gives me is "To obtain the solution explicitly we must solve for y in terms of x. This is a simple matter in this case since the equation is quadratic in y" and then they jump to the solution:
y = 1 +/- sqrt(x^3 + 2x^2 +2x + 4)

 

[dav2008]

Well you said it yourself: it's a quadratic in terms of y.

Do you know how to solve quadratic equations?

 

[davegillmour]

Well you said it yourself: it's a quadratic in terms of y.

Do you know how to solve quadratic equations?

So would the expression x^3 + 2x^2 +2x + 3 be treated as the 'c' value in the quadratic eq?

 

[dav2008]

Yep.

Remember that the quadratic equation let's you solve for a certain variable (we'll call it w here) when you have an equation in the form of aw^2+bw+c=0

 

[davegillmour]

Alright I got it, thanks a lot

 

[HallsofIvy]

So would the expression x^3 + 2x^2 +2x + 3 be treated as the 'c' value in the quadratic eq?

Actually, no. Since your equation is y^2 - 2y = x^3 + 2x^2 +2x + 3
and the quadratic is normally written ay^2+ by+ c= 0,
c= -x^3- 2x^2- 2x- 3.