Example 16 (1999 Canadian Mathematical Olympiad) Find all real solutions to the equation $4 x^{2}-40[x]+51=0$.

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Solution 1 Since $[x] \leqslant x$, we have
$0=4 x^{2}-40[x]+51 \geqslant 4 x^{2}-40 x+51=(2 x-3)(2 x-17)$,
which means $\frac{3}{2} \leqslant x \leqslant \frac{17}{2}$.
Thus, $[x]=1,2,3,4,5,6,7,8$.
When $[x]=1$, the equation becomes $4 x^{2}+11=0$, which has no real solution.
When $[x]=2$, the equation becomes $4 x^{2}-29=0$,
and by $\frac{3}{2} \leqslant x \leqslant \frac{7}{2}$ we get $x=\frac{\sqrt{29}}{2}$.
When $[x]=3$, we get $x=\frac{\sqrt{69}}{2}$, which contradicts $[x]=3$;
When $[x]=4$, we get $x=\frac{\sqrt{109}}{2}$, which contradicts $[x]=4$;
When $[x]=5$, we get $x=\frac{\sqrt{149}}{2}$, which contradicts $[x]=5$;
When $[x]=6$, we get $x=\frac{\sqrt{189}}{2}$, and at this time $\left[\frac{\sqrt{189}}{2}\right]=6$, so $x=\frac{\sqrt{189}}{2}$ is a solution to the equation; when $[x]=7$, we get $x=\frac{\sqrt{229}}{2}$, and at this time $\left[\frac{\sqrt{229}}{2}\right]=7$, so $x=\frac{\sqrt{229}}{2}$ is a solution to the equation; when $[x]=8$, we get $x=\frac{\sqrt{269}}{2}$, and at this time $\left[\frac{\sqrt{269}}{2}\right]=8$, so $x=\frac{\sqrt{269}}{2}$ is a solution to the equation. From the above, the solution set of the given equation is $\left\{\frac{\sqrt{29}}{2}, \frac{\sqrt{189}}{2}, \frac{\sqrt{229}}{2}, \frac{\sqrt{269}}{2}\right\}$.
Solution 2 By property (3), we know $[x] \leqslant x0, \\
4[x]^{2}-4[x]+51 \leqslant 0,
\end{array}\right.
$

which means $\left\{\begin{array}{l}(2[x]-5)(2[x]-11)>0, \\ (2[x]-3)(2[x]-7) \leqslant 0,\end{array}\right.$

thus $[x]=2$ or $[x]=6$ or $[x]=7$ or $[x]=8$, substituting these into the equation we get $x=\frac{\sqrt{29}}{2}$, or $x=$ $\frac{\sqrt{189}}{2}$, or $x=\frac{\sqrt{229}}{2}$, or $x=\frac{\sqrt{269}}{2}$. After verification, these 4 values are all solutions to the original equation.
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