[Example 5.5.1] Let $a$ be a negative integer, find the value of $a$ such that the equation
$$
a x^{2}-2(a-3) x+(a-2)=0
$$

has at least one integer solution $x$.
The positive solution to this problem is:
Since $a \neq 0$, the solutions to the equation are
$$
x=\frac{a-3 \pm \sqrt{9-4 a}}{a}.
$$

To ensure that at least one of the solutions is an integer, a necessary condition is that $9-4 a$ must be a perfect square. There are many solutions for negative integer $a$ that make $9-4 a$ a perfect square, such as $a=-4$, $-10$, $-18$, $-28$, $-40$, etc. After $9-4 a$ is a perfect square, it is still necessary to verify whether $x$ is an integer, which can be difficult to calculate.

Since the positive approach is difficult, we can think from the opposite side. Instead of solving for $x$, we solve for $a$, utilizing the characteristic that $a$ is a negative integer to handle the problem.
At this point, the given equation becomes
$$
\left(x^{2}-2 x+1\right) a+6 x-2=0.
$$
When $x=1$, the above equation does not hold, so $x \neq 1$, and $(x-1)^{2} \neq 0$,
$$
a=\frac{-6 x+2}{x^{2}-2 x+1} \leqslant-1
$$

Solving the above inequality,
$$
\begin{array}{c}
-6 x+2 \leqslant-x^{2}+2 x-1, \\
x^{2}-8 x+3 \leqslant 0, \\
4-\sqrt{13} \leqslant x \leqslant 4+\sqrt{13} \quad \text { and } x \neq 1 .
\end{array}
$$

Thus, the possible integer values for $x$ are $2, 3, 4, 5, 6, 7$.
The next step is to substitute $x=2, 3, 4, 5, 6, 7$ into the original equation to check which values of $x$ make $a$ a negative integer.
Upon inspection, $x=4, 5, 6, 7$ do not yield integer values for $a$.
When $x=2$, $a=-10$, and the equation becomes
$$
\begin{array}{c}
5 x^{2}-13 x+6=0, \\
x=2, x=\frac{3}{5}.
\end{array}
$$

When $x=3$, $a=-4$, and the equation becomes
$$
\begin{array}{c}
2 x^{2}-7 x+3=0, \\
(x-3)(2 x+1)=0, \\
x=3, x=\frac{1}{2}.
\end{array}
$$

Therefore, the values of $a$ are $a=-4, -10$.