Example 3 Let $m$ be an integer, and $4<m<40$, and 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: From $\Delta=[-2(2 m-3)]^{2}-4\left(4 m^{2}-\right.$ $14 m+8)=4(2 m+1)$ and $4<m<40$,
we have $9<2 m+1<81$.
Since $2 m+1$ is an odd number,
thus $2 m+1=5^{2}$ or $7^{2}$.
Then $m=12$ or 24.
When $m=12$, $x^{2}-42 x+416=0$, $x_{1}=26, x_{2}=16$;
When $m=24$, $x^{2}-90 x+1976=0$, $x_{1}=52, x_{2}=38$.