Help with common form

Hey I was wondering if I could have some help with a certain problem that has been bugging me. I would really like an explination than a direct answer, since I want to be able to comprehend the idea and not just get an easy answer. Hope you can help.

ax^2 = x^2 - x

And I need to find for what values of a does the equation have: no solutions, exactly one solution, and two solutions.

So I'm stuck from here on out, I've tried a lot of different things but basically I think I'm starting the problem wrong. First I need to get it in common form, correct? Can I get some help with this part?

Then I can take it from there, I know to get 1 solution a=0, 2 a>0, and none a<0 correct? And I just use the quadratic formula  part to find the (radical right? get my terms mixed up) b^2-4(a)(c).

So basically, help me get start the problem/ get it in common form and I can take it from there. Thanks a lot.

\frac{-b \pm \sqrt {b^2-4ac}}{2a}
Plugging in your values a is 1-a, b is -1, c is 0 so:
\frac{1 \pm \sqrt {-1^2-4(1-a)0}}{2(1-a)}
\frac{1 \pm \sqrt {1}}{2(1-a)}
\frac{1 \pm 1}{2(1-a)}
Getting rid of the ±:
\frac{0}{2(1-a)} = 0
\frac{2}{2(1-a)} = \frac{1}{1-a}