Solving for x, but only getting half

You cannot divide by x if it is zero. So your method assumes that x is not zero already, and obviously then you can't get it as a solution! In solving such problems, never divide by a quantity unless (1) you are sure it is nonzero or (2) you split out a separate case for when it is zero.

 

Maybe ##x^2-2x = x(x-2)## will help you? There are other ways as well. Such as ##x^2=2x##. Note that for the 2nd equation 0 is a solution (##0^2 = 2*0##). Then we can proceed to find non-zero solutions.

 

##x^2=2x## implies that either ##x=0## or ##x=2## (because if ##x\neq 0##, we can divide by ##x## to get ##x=2##).

The above doesn't prove that the solutions are 0 and 2. It just proves that no number other than 0 or 2 can be a solution. It is however easy to verify that both 0 and 2 are solutions: Since ##0^2=0=2\cdot 0##, we know that ##0## is a solution. Since ##2^2=2\cdot 2##, we know that ##2## is a solution.

This is how I think when I solve these problems. I like this method better than both factorization and the formula for solutions.

An alternative to explicitly verifying that 2 is a solution is to note that if ##x\neq 0##, the statements ##x^2-2x=0##, ##x^2=2x## and ##x=2## are all equivalent.

In case it's not perfectly clear what's going on in the first step: We already knew without looking at the equation that either ##x=0## or ##x\neq 0##. So we consider these two possibilities separately. If ##x=0##, the equation tells us nothing. (OK, it tells us that ##0=0##, but we already knew that). If ##x\neq 0##, then the equation tells us that ##x=2##. We have now ruled out all non-zero solutions other than ##2##.