Solutions to $[x^2]+2[x]=3x\text{ where } 0\le x\le 2$ 
Find all solutions to $$[x^2]+2[x]=3x\text{ where } 0\le x\le 2$$
  and $[x]=\lfloor x\rfloor$

$$$$
I managed to simplify this to $[x^2]-[x]=3\{x\}$. Thus, $$[x^2]-[x]=\{0,1,2\}$$ However I got stuck here and was unable to proceed further. Somehow I have a feeling that there's a much neater approach than what I've taken.$$$$
Any help would be greatly appreciated. Many thanks!

If  $0\leq x<1$, $LHD=0$.  
$1\leq x<\sqrt2$, $LHD=3$
$\sqrt2\leq x<\sqrt3,LHD=4$ 
$\sqrt3\leq x<2,LHD=5$
So x is a rational number by feature of equation, $x=0,1,2$
