1985 AMC 12 Problem 30

Set $n=\lfloor x\rfloor$ and solve the equation for $x^2$

Enforce $n\le x\lt n+1$ to determine which integer values of $n$ work

Let $n=\lfloor x\rfloor.$ A solution must be positive and satisfy $$\begin{aligned} x&=\sqrt{10n-\frac{51}{4}},\\ n&\le x\lt n+1. \end{aligned}$$ The lower inequality is equivalent to $$4n^2-40n+51\le0,$$ so $n=2,3,\ldots,8.$ The upper inequality is equivalent to $$4n^2-32n+55\gt0.$$ Among $2,\ldots,8,$ this holds exactly for $n=2,6,7,8.$ Each interval contributes one root, so there are four real solutions.

Therefore the correct answer is E.