4・30 Find all solutions to the equation $x^{2}-8[x]+7=0$. Here, $[x]$ denotes the greatest integer less than or equal to $x$.

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4・30 Find all solutions to the equation $x^{2}-8[x]+7=0$. Here, $[x]$ denotes the greatest integer less than or equal to $x$.
[Solution] Let $x$ be the root of the given equation, and $[x]=n$. From the original equation, we have
$$x^{2}+7=8 n$$

This indicates that $n>0$. Using $n \leqslant x < n+1$, we get
$$\left\{\begin{array}{l}
n^{2}+7<8(n+1) \\
n^{2}-8 n+7 \leqslant 0
\end{array}\right.$$

Solving this, we get $\quad 1 \leqslant n<2$ 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 $x=7$.