<s>[INST] If $x\in\mathbb R$, solve $4x^2-40\lfloor x\rfloor+51=0$. 
If $x\in\mathbb R$, solve $$4x^2-40\lfloor x\rfloor+51=0$$
  where $\lfloor x\rfloor$ denotes the integer part of the number. $\lfloor x\rfloor\le x$ and $\lfloor x\rfloor=x-\{x\}$, where $\{x\}$ marks the fraction part of the number. $0\le\{x\}<1$.

Not sure if all these denotions are conventions that always mean what I've mentioned.
Also, I'm not too sure if $x\in\mathbb R$, since the problem doesn't mention it, but it seems quite obvious that it most likely is that way.
 [/INST] Since $$x-1\lt \lfloor x\rfloor \le x,$$
we have
$$\begin{align}x-1\lt \frac{4x^2+51}{40}\le x&\iff 4x^2-40x+91\gt0\ \text{and}\ 4x^2-40x+51\le 0\\&\iff 1.5\le x\lt 3.5\ \text{or}\ 6.5\lt x\le 8.5.\end{align}$$
1) When $1.5\le x\lt 2\Rightarrow \lfloor x\rfloor=1$, 
$$4x^2-40\times 1+51=0$$
does not have any real solution.
2) When $2\le x\lt 3\Rightarrow \lfloor x\rfloor=2$, 
$$4x^2-40\times 2+51=0\Rightarrow x=\pm \sqrt{29}/2\Rightarrow x=\sqrt{29}/2.$$
3) When $3\le x\lt 3.5\Rightarrow \lfloor x\rfloor=3$, 
$$4x^2-40\times 3+51=0\Rightarrow x=\pm \sqrt{69}/2.$$
But these don't satisfy $3\le x\lt 3.5.$
4) When $6.5\lt x\lt 7\Rightarrow \lfloor x\rfloor=6$, 
$$4x^2-40\times 6+51=0\Rightarrow x=\pm 3\sqrt{21}/2\Rightarrow x=3\sqrt{21}/2.$$
5) When $7\le x\lt 8\Rightarrow \lfloor x\rfloor=7$, 
$$4x^2-40\times 7+51=0\Rightarrow x=\pm \sqrt{229}/2\Rightarrow x=\sqrt{229}/2.$$
6) When $8\le x\le 8.5\Rightarrow \lfloor x\rfloor=8$, 
$$4x^2-40\times 8+51=0\Rightarrow x=\pm \sqrt{269}/2\Rightarrow x=\sqrt{269}/2.$$
Hence, the answer is 
$$x=\frac{\sqrt{29}}{2},\frac{3\sqrt{21}}{2},\frac{\sqrt{229}}{2},\frac{\sqrt{269}}{2}.$$
</s>