9. The solutions to the equation $x^{2}-8[x]+7=0$ are

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9. $x=1, \sqrt{33}, \sqrt{41}, 7$

Let $[x]=n$, then the known equation can be written as $x^{2}+7=8 n$, so $n>0$.
By $n \leqslant x0$.
Solving, we get $1 \leqslant n<2$ or $4<n \leqslant 7$, hence $n=1,5,6,7$. Then we have $x^{2}+7=8,40,48,56$ solving for $x= \pm 1$, $\pm \sqrt{33}, \pm \sqrt{41}, \pm 7$

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