Solve: $[x]^2 = x + 2\{x\}$, where [.] and {.} denote the greatest integer and the fractional part functions, respectively.

$\{x\}=\frac{2}{3},0,0,\frac{2}{3}$

$x=-\frac{1}{3},0,1,\frac{8}{3}$

$[x]^2 = x + 2\{x\}⇒[x]^2 = [x] + 3\{x\}$

so $\{x\} = \frac{[x]^2 -[x]}{3}$

so $0≤\frac{[x]^2 -[x]}{3}<1$

so $[x]^2 -[x]>0$   $[x]^2 -[x]<3$

so $[x]([x]-1)>0$

$[x]∈(-∞,0)∪[1,∞)$

$[x]^2 -[x]-3=0⇒[x]=\frac{1±\sqrt{13}}{2}$ so for $[x]^2 -[x]-3<0$

$⇒[x]∈[\frac{1-\sqrt{13}}{2},0]∪[1,\frac{1+\sqrt{13}}{2}]$

$⇒[x]=-1,0,1,2$

$\{x\}=\frac{[x]^2 -[x]}{3}⇒\{x\}=\frac{2}{3},0,0,0$

so $x=[x]^2 -2\{x\}⇒x=-\frac{1}{3},0,1,\frac{8}{3}$