$\begin{array}{c}\text { For example, solve the equation } \\ 4 x^{2}-40 [ x ] \\ +51=0 .\end{array}$
Let $x=[x]+r, \quad-$ $0 \leqslant r<1$.

Then we have $4 r^{2}+8 r\lceil x\rfloor$ $+4[x]^{2}-40[x]+51=0$,
Clearly, $[x] \geqslant \frac{51}{40}$, and since $r \geqslant 0$, we have
$$
0 \leqslant \frac{-2[x]+\sqrt{40[x]-51}}{2}<1 .
$$

Solving this inequality, noting that $[x]$ is an integer, we get
$$
[x]=2,6,7,8 \text {. }
$$

Thus, we can find four roots
$$
\begin{array}{l}
x_{1}=\frac{\sqrt{29}}{2}, x_{2}=3 \sqrt{21}, \\
x_{3}=\sqrt{ } \frac{229}{2}, x_{4}=\sqrt{269} .
\end{array}
$$