Help needed, rearranging polynomial for inverse equation

Hi, I need to rearrange an equation:

y = ax^2 + bx + c

to the form of:

x = ?

I'm not entirely sure how to go about this and the examples I've found require the equation to be in a different form. Any tips or a point in the right direction would be great!

 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Complete the square.
 Great thanks! That in mind I've got: x = $\frac{\sqrt{y - c - \frac{b^{2}}{4a}} - \frac{b}{2\sqrt{a}}}{\sqrt{a}}$

Help needed, rearranging polynomial for inverse equation

Check out the Wolfram Equation Solver:

http://www.wolframalpha.com/examples...onSolving.html

You would want to "solve an equation with parameters". Their answer looks a bit different, so you can look at their step by step breakdown and see whether your answer is equivalent.

 Blog Entries: 5 Recognitions: Homework Help Science Advisor Do you know the quadratic formula, which is the default solution of ax² + bx + c = 0? Or is that what you are trying to prove here? Because if not, you can pull y to the other side of the equals sign and apply the quadratic formula.
 Recognitions: Gold Member Science Advisor Staff Emeritus Complete the square or use the quadratic formula. Most people learn how to solve quadratic equations before they learn about "inverse functions". Also, at some point you will have a "plus or minus". Unless your domain is restricted, a quadratic function will NOT have an inverse function.
 I'm actually writing a program that works out a, b and c, but then needs to work out x given y. I probably used the wrong terminology to describe something along the way ^^ The answer I first wrote was generated by getting the equation in the form of: y = (dx + e)^2 + f and then working out d, e and f. The wolfram example is much nicer solution though, and more efficient computer wise :) Thanks a lot for the help!
 Blog Entries: 5 Recognitions: Homework Help Science Advisor If you already have a square you can solve for p = (dx + e) first.