$14 \cdot 16$ Solve the equation $x^{2}-2 x-3=12 \cdot\left[\frac{x-1}{2}\right]$.
(China Sichuan Province Junior High School Mathematics League, 1990)
\[ \text{Solution: The original equation can be transformed into} \]
\[
\begin{array}{c}
\left(\frac{x-1}{2}\right)^{2}-1=3 \cdot\left[\frac{x-1}{2}\right] . \\
\text{Let } \left[\frac{x-1}{2}\right]=m, \frac{x-1}{2}=m+r, \quad 0 \leqslant r<1 .
\end{array}
\]

From (1), we know that \( m \) is a non-negative integer, and \( \left(\frac{x-1}{2}\right)^{2} \) is an integer.
\[
\left(\frac{x-1}{2}\right)^{2}=m^{2}+2 m r+r^{2} \text{. }
\]

Let \( 2 m r+r^{2}=n \), then \( n \) is a non-negative integer.
Substituting (2) into (1) gives \( m^{2}+2 m r+r^{2}-1=3 m \),
\[
m^{2}-3 m+n-1=0 \text{. }
\]

So, \( m=\frac{3 \pm \sqrt{9-4(n-1)}}{2} \).
When \( n=1 \), \( m=0,3 \);
When \( n=3 \), \( m=1,2 \).
For other non-negative integers \( n \), \( m \) is not a non-negative integer.
When \( m=0 \), \( r^{2}=1 \), which contradicts \( 0 \leqslant r<1 \).
When \( m=1 \),
\[
r^{2}+2 r-3=0
\]

Solving gives \( r=1 \) or -3, which contradicts \( 0 \leqslant r<1 \).
When \( m=2 \), from
\[
r^{2}+4 r-3=0
\]

we get \( r=-2+\sqrt{7} \), thus
\[
\begin{array}{c}
\frac{x-1}{2}=m+r=2+(-2+\sqrt{7})=\sqrt{7}, \\
x=1+2 \sqrt{7} .
\end{array}
\]

When \( m=3 \), from
\[
r^{2}+6 r-1=0
\]

we get \( r=-3+\sqrt{10} \), thus
\[
\begin{array}{c}
\frac{x-1}{2}=3+(-3+\sqrt{10})=\sqrt{10}, \\
x=1+2 \sqrt{10} .
\end{array}
\]

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