If $X$ is uniform on the interval $[0, 2]$ , find the PDF of $X^2-2X$ If $X$ is uniform on the interval $[0, 2]$ find the PDF of $Y =
X^2-2X$. 
I solved for $X$ and got $x=1 + \sqrt{1+y}$ or $x=1 - \sqrt{1+y}$. Not sure what to do from here.
 A: Note that $Y=(X-1)^2-1$. Now let $W=X-1$. It is nearly obvious that $W$ is uniform on $[-1,1]$. 
We find the distribution of $W^2$. Since $Y=W^2-1$, it is then easy to find the distribution of $Y$.
Let $Z=W^2$. Then (for $0\le z\le 1$), 
$$\Pr(Z\le z)=\Pr(W^2\le z)=\Pr(|W|\le \sqrt{z})=\frac{2\sqrt{z}}{2}=\sqrt{z}.$$ 
Remark: The problem can also be solved using a variant of your calculation. We have $Y\le y$ if and only if $X^2-2X\le y$. This inequality holds if and only if
$$1-\sqrt{1+y}\le X\le 1+\sqrt{1+y}.$$
If $y\ge 0$, the above probability is $1$. So the cdf of $Y$ is $1$ for $y \ge 0$.  
If $y\lt -1$, the above condition is an impossible one, so the probability is $0$. Thus the cdf of $Y$ is $0$ when $y\lt -1$. So the only interesting $y$ are between $-1$ and $0$. 
For $-1\le y\le 0$, the length of the interval $[1-\sqrt{1+y},1+\sqrt{1+y}]$ is $2\sqrt{1+y}$, and the interval lies within $[0,2]$. So the probability we land in that interval is $\frac{2\sqrt{1+y}}{2}$, that is, $\sqrt{1+y}$.
Now that we have the cdf of $Y$, we can differentiate to get the density. 
