Problem 761

8. $x=1+2 \sqrt{7}$ or $x=1+2 \sqrt{10}$.

Let $\frac{x-1}{2}=y$, then $x^{2}-2 x-3=(x-1)^{2}-4=4 y^{2}-4$, i.e., $4 y^{2}-4=12[y]$, which means $y^{2}-1=3[y]$. Therefore, $y^{2}-1=3[y] \leqslant 3 y$, and $y^{2}-1=3[y]>3(y-1)$. From $y^{2}-1 \leqslant 3 y$ we get $\frac{3-\sqrt{13}}{2} \leqslant y \leqslant \frac{3+\sqrt{13}}{2}$, and from $y^{2}-1>3(y-1)$ we get $y>2$ or $y<1$, so $\frac{3-\sqrt{13}}{2} \leqslant y<1$ or $2<y \leqslant \frac{3+\sqrt{13}}{2}$. Therefore, $[y]=-1$, or 0, or 2, or 3.

If $[y]=-1$, then from $y^{2}-1=3[y]$ we get $y^{2}=-2$, which is a contradiction! If $[y]=0$, then from $y^{2}-1=3[y]$ we get $y = \pm 1$, which is a contradiction! If $[y]=2$, then $y=\sqrt{7}$, if $[y]=3$, then $y=\sqrt{10}$, in summary, $y=\sqrt{7}$ or $\sqrt{10}$.
Therefore, $x=1+2 \sqrt{7}$ or $x=1+2 \sqrt{10}$.