(solved by Victor Quach). Find all real solutions of the equation:

$$
4 x^{2}-40[x]+51=0
$$

where $[x]$ denotes the integer part of $x$.
. Note that

$$
(2 x-3)(2 x-17)=4 x^{2}-40 x+51 \leqslant 4 x^{2}-40[x]+51=0
$$

which gives $1.5 \leqslant x \leqslant 8.5$ and $1 \leqslant[x] \leqslant 8$. From the equation,

$$
x=\frac{\sqrt{40[x]-51}}{2}
$$

and therefore it is necessary that

$$
[x]=\left[\frac{\sqrt{40[x]-51}}{2}\right]
$$

By testing $[x]=1,2, \ldots, 8$ in this equation, we find that $[x]$ can only be equal to 2, 6, 7, or 8. We deduce that the only possible solutions for $x$ are $\sqrt{29} / 2, \sqrt{189} / 2, \sqrt{229} / 2$ and $\sqrt{269} / 2$. A quick check confirms that these values work.