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 $[x]=\lfloor x\rfloor$

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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!

Write
$$x={k\over3}\qquad(0\leq k\leq 6)$$
with integer $k$. The condition then is
$$\left\lfloor{k^2\over 9}\right\rfloor+2\left\lfloor{k\over 3}\right\rfloor=k\ .$$
Within the given range only $k=0$ and $k=3$ do qualify, so that we obtain the two solutions $x_1=0$ and $x_2=1$.
