imc

To further grasp at this equation, we rearrange the equation into $$\lfloor x{\rfloor }^2=3x-2.$$ Thus, $3x-2$ is a perfect square and nonnegative. It is now much more apparent that $x\ge 2/3,$ and that $x=2/3$ is a solution. Additionally, by observing the RHS, $x<4,$ as $$\lfloor 4{\rfloor }^2>3\cdot 4,$$ since squares grow quicker than linear functions. Now that we have narrowed down our search, we can simply test for intervals $[2/3,1],[1,2],[2,3],[3,4).$ This intuition to use intervals stems from the fact that $x=1,2$ are observable integral solutions. Notice how there is only one solution per interval, as $3x-2$ increases while $\lfloor x{\rfloor }^2$ stays the same. Finally, we see that $x=3$ does not work, however, through setting $\lfloor x{\rfloor }^2=9,$ $x=11/3$ is a solution and within our domain of $[3,4).$ This provides us with solutions $\left(\frac{2}{3},1,2,\frac{11}{3}\right),$ thus the final answer is $\boxed{4}.$