How many distinct values of $x$ satisfy
$\lfloor{x}\rfloor^2-3x+2=0$, where $\lfloor{x}\rfloor$ denotes the largest integer less than or equal to $x$?
$\textbf{(A) } \text{an infinite number} \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 0$
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 3\cdot4,\]
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 $\left(\frac23, 1, 2, \frac{11}{3}\right),$ thus the final answer is $\boxed{(\text{B}) \ 4}.$
~mathbrek, happyhari