$4 \cdot 30$ Solve the equation $x^{2}-8[x]+7=0$. Here $[x]$ denotes the greatest integer less than or equal to $x$.

Find all solutions to the equation.
[Solution] Let $x$ be the root of the given equation, and $[x]=n$. From the original equation, we have
$$
x^{2}+7=8 n \text {. }
$$

This indicates that $n>0$. Using $n \leqslant x < n+1$, we get
$$
\left\{\begin{array}{l}
n^{2}+7 \leqslant 8 n, \\
n^{2}+2 n+8>8 n .
\end{array}\right.
$$
Solving these, we get $1 \leqslant n < 2$ or $4 < n \leqslant 7$.
Thus, $n=1,5,6,7$.
For $n=1$, we get $x=1$,
For $n=5$, we get $x=\sqrt{33}$,
For $n=6$, we get $x=\sqrt{41}$,
For $n=7$, we get $x=7$.