Example 4 If the fractional parts of $9+\sqrt{13}$ and $9-\sqrt{13}$ are $a$ and $b$ respectively, then $a b-4 a+3 b-2=$ $\qquad$
Solution: $\because 3<\sqrt{13}<4$,
$\therefore 9+\sqrt{13}$ has an integer part of 12, and a decimal part
$$
\begin{array}{l} 
a=\sqrt{13}-3 . \\
\because-4<-\sqrt{13}<-3,
\end{array}
$$

i.e., $0<4-\sqrt{13}<1$,
$\therefore 9-\sqrt{13}$ has an integer part of 5, and a decimal part $b$ $=4-\sqrt{13}$.
Thus, $a b-4 a+3 b-2$
$$
\begin{array}{l}
=(a+3)(b-4)+10 \\
=\sqrt{13} \times(-\sqrt{13})+10=-3 .
\end{array}
$$