$14 \cdot 72$ Find all real solutions of the equation $4 x^{2}-40[x]+51=0$.
(Canadian Mathematical Olympiad, 1999)
[Solution] Since $[x] \leqslant x$,
we have $0=4 x^{2}-40[x]+51 \geqslant 4 x^{2}-40 x+51$
$$
=(2 x-3)(2 x-17) \text {. }
$$

Thus, $\frac{3}{2} \leqslant x \leqslant \frac{17}{2}$.
Therefore, $[x]=1,2,3,4,5,6,7,8$.
When $[x]=1$, the equation becomes $4 x^{2}+11=0$ which has no real solutions.
When $[x]=2$, the equation becomes $4 x^{2}-29=0$,
and by $\frac{3}{2} \leqslant x \leqslant \frac{7}{2}$ we get $\quad x=\frac{\sqrt{29}}{2}$.
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 $\left[\frac{\sqrt{189}}{2}\right]=6$,
Therefore, $x=\frac{\sqrt{189}}{2}$ is a solution of the equation.
When $[x]=7$, we get $x=\frac{\sqrt{229}}{2}$, and at this time $\left[\frac{\sqrt{229}}{2}\right]=7$,

Therefore, $x=\frac{\sqrt{229}}{2}$ is a solution of the equation.
When $[x]=8$, we get $x=\frac{\sqrt{269}}{2}$, and at this time $\left[\frac{\sqrt{269}}{2}\right]=8$,
Therefore, $x=\frac{\sqrt{269}}{2}$ is a solution of the equation.
From the above, the solution set of the given equation is $\left\{\frac{\sqrt{29}}{2}, \frac{\sqrt{189}}{2}, \frac{\sqrt{229}}{2}, \frac{\sqrt{269}}{2}\right\}$.