smc

We rearrange the equation as $4x^2=40\lfloor x\rfloor -51$, where the right-hand side is now clearly an integer, meaning that $4x^2=n$ for some non-negative integer $n$. Therefore, in the case where $x\ge 0$, substituting $x=\frac{\sqrt{n}}{2}$ gives $$40\lfloor \frac{\sqrt{n}}{2}\rfloor -51=n.$$ To proceed, let $a$ be the unique non-negative integer such that $a\le \frac{\sqrt{n}}{2}<a+1$, so that $$\begin{array}{rl}& \lfloor \frac{\sqrt{n}}{2}\rfloor =a,\text{ and}\\ & 4a^2\le n<4a^2+8a+4,\end{array}$$ and our equation reduces to $$40a-51=n.$$ The above inequalities therefore become $$4a^2\le 40a-51<4a^2+8a+4⟺4a^2-40a+51<0\text{ and }4a^2-32a+55>0,$$ where the first inequality can now be rewritten as $(2a-10{)}^2\le 49$, i.e. $\left|2a-10\right|\le 7$. Since $(2a-10)$ is even for all integers $a$, we must in fact have $$\begin{array}{rl}\left|2a-10\right|\le 6& ⟺\left|a-5\right|\le 3\\ & ⟺2\le a\le 8.\end{array}$$ The second inequality similarly simplifies to $(2a-8{)}^2>9$, i.e. $\left|2a-8\right|>3$. As $(2a-8)$ is even, this is equivalent to $$\begin{array}{rl}\left|2a-8\right|\ge 4& ⟺\left|a-4\right|\ge 2\\ & ⟺a\ge 6\text{ or }a\le 2,\end{array}$$ so the values of $a$ satisfying both inequalities are $2$, $6$, $7$, and $8$. Since $n=40a-51$, each of these distinct values of $a$ gives a distinct solution for $n$, and thus for $x=\frac{\sqrt{n}}{2}$, giving a total of $4$ solutions in the $x\ge 0$ case. As $4$ is already the largest of the answer choices, this suffices to show that the answer is $\text{(E)}$, but for completeness, we will show that the $x<0$ case indeed gives no other solutions. If $x=-\frac{\sqrt{n}}{2}$ (and so $n>0$), we require $$40\lfloor -\frac{\sqrt{n}}{2}\rfloor -51=n,$$ and recalling that $\lfloor -x\rfloor =-\lceil x\rceil$ for all $x$, this equation can be rewritten as $$-40\lceil \frac{\sqrt{n}}{2}\rceil -51=n.$$ Since $n$ is positive, the least possible value of $\lceil \frac{\sqrt{n}}{2}\rceil$ is $1$, but this means $$\begin{array}{rl}n& =-40\lceil \frac{\sqrt{n}}{2}\rceil -51\\ & \le -40\cdot 1-51\\ & =-91,\end{array}$$ which is a contradiction. Therefore the $x<0$ case indeed gives no further solutions, confirming that the total number of solutions is precisely $\boxed{4}$.