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The solution should say:

Let $\displaystyle  t=\left\lfloor\frac{x-2}{3} \right\rfloor$.  Then $\displaystyle x=\frac{4t+3}{2}$.  By definition, we have $\displaystyle t\leq\frac{x-2}{3} < t+1$, thus $\displaystyle t\leq\frac{4t-1}{6} < t+1$.  Solving for $t$, we get $\displaystyle -\dfrac{7}{2} < t \leq - \dfrac{1}{2}$.  So $t=-1,-2,-3$. Finally, plug $t$ back in to solve for $x$.

The solutions are $-\dfrac{1}{2}$, $-\dfrac{5}{2}$, and $-\dfrac{9}{2}$.

The solution has been updated now. Thanks for letting us know!