Solution: The original equation can be rewritten as
$$\left(\frac{x-1}{2}\right)^{2}-1=3 \cdot\left[\frac{x-1}{2}\right] \text {. }$$

From $\left[\frac{x-1}{2}\right]$ being an integer and equation (1), we know that $\left(\frac{x-1}{2}\right)^{2}$ is also an integer.
Let $\left[\frac{x-1}{2}\right]=m$, then
$$\frac{x-1}{2}=m+r, 0 \leqslant r<1 \text {. }$$

From equation (1), we know that $m$ is a non-negative integer, and from $\left(\frac{x-1}{2}\right)^{2}$ being an integer and
$$\left(\frac{x-1}{2}\right)^{2}=m^{2}+2 m r+r^{2}$$

we know that $2 m r+r^{2}$ is also an integer.
Let $2 m r+r^{2}=n$, then $n$ is a non-negative integer.
Substituting equation (2) into equation (1) yields
$m^{2}-3 m+n-1=0$.
$$m=\frac{3 \pm \sqrt{9-4(n-1)}}{2}=\frac{3 \pm \sqrt{13-4 n}}{2} \text {. }$$

From $13-4 n \geqslant 0$ we know
$0 \leqslant n \leqslant \frac{13}{4}, n=0,1,2,3$.
When $n=0,2$, $m$ is not an integer;
When $n=1$, $m=0$ or 3;
When $n=3$, $m=1$ or 2.
(i) When $m=0, n=1$, $r^{2}=1$.

This contradicts $0 \leqslant r<1$.
(ii) When $m=1, n=3$, we have $r^{2}+2 r-3=0, r=1$ or -3.

This contradicts $0 \leqslant r<1$.
(iii) When $m=2, n=3$, we have
$$r^{2}+4 r-3=0 \text {. }$$

Solving, we get $r=-2+\sqrt{7}$.
$$\begin{array}{l} \therefore \frac{x-1}{2}=m+r=2+(-2+\sqrt{7})=\sqrt{7} . \\ x=1+2 \sqrt{7} . \end{array}$$
(iv) When $m=3, n=1$, we have
$$r^{2}+6 r-1=0 \text {. }$$

Solving, we get $r=-3+\sqrt{10}$.
Thus, $\frac{x-1}{2}=3+(-3+\sqrt{10})=\sqrt{10}$.
$$x=1+2 \sqrt{10} \text {. }$$

Therefore, the solutions to the original equation are
$$x=1+2 \sqrt{7} \text { or } x=1+2 \sqrt{10} \text {. }$$