Example 5 Let $m \in \mathbf{Z}$, and $4<m<40$, the equation $x^{2}-2(2 m-3) x+4 m^{2}-14 m+8=0$ has two integer roots. Find the value of $m$ and the roots of the equation.
Solution: Since the equation has integer roots, then
$$
\begin{aligned}
\Delta & =[-2(2 m-3)]^{2}-4\left(4 m^{2}-14 m+8\right) \\
& =4(2 m+1)
\end{aligned}
$$

is a perfect square.
Thus, $2 m+1$ is a perfect square.
Also, since $m \in \mathbf{Z}$ and $4<m<40$,
it follows that $2 m+1$ is a perfect square only when $m=12$ or 24.
Since $x=(2 m-3) \pm \sqrt{2 m}+1$, we have:
when $m=12$, $x_{1}=16, x_{2}=26$;
when $m=24$, $x_{3}=38, x_{4}=52$.
(2) First, use $\Delta \geqslant 0$ to find the range of the parameter.