$14 \cdot 14$ Find all solutions to the equation $x^{2}-8[x]+7=0$.
(13th All-Russian Mathematical Olympiad, 1987)
[Solution] Let $x$ be the root of the given equation. And let $[x]=n$.
Thus, the given equation becomes
$$
x^{2}+7=8 n \text {. }
$$

Therefore, $n>0$. From

we get
$$
\begin{array}{l}
n \leqslant x < n+1.
\end{array}
$$

Solving this, we get $1 \leqslant n<2$ or $4<n \leqslant 7$.
That is, $n=1,5,6,7$.
Then we have
$$
x^{2}+7=8,40,48,56 \text {. }
$$

Solving this, we get
$$
x= \pm 1, \pm \sqrt{33}, \pm \sqrt{41}, \pm 7 \text {. }
$$

Since $n \geqslant 0$, we have $x \geqslant 0$, so only
$$
x=1, \sqrt{33}, \sqrt{41}, 7 \text {. }
$$