Example 5 In the sequence $\left\{a_{n}\right\}$, $a_{1}=2$, and $a_{n+1}=\frac{2 a_{n}-4}{a_{n}+6}\left(n \in \mathbf{N}^{*}\right)$, find its general term formula $a_{n}$.
Solution: The characteristic equation is $x=\frac{2 x-4}{x+6}$. Solving it yields $x_{1}=x_{2}=-2$.
Let $a_{n}=b_{n}-2\left(n \in \mathbf{N}^{*}\right)$, substituting it into the original recurrence relation gives $b_{n+1}=\frac{4 b_{n}}{b_{n}+4}\left(n \in \mathbf{N}^{*}\right)$.
Since $b_{1}=a_{1}+2=4 \neq 0$, we have $b_{k} \neq 0\left(k \in \mathbf{N}^{*}\right)$.
Thus, $\frac{1}{b_{n+1}}=\frac{1}{b_{n}}+\frac{1}{4}\left(n \in \mathbf{N}^{*}\right)$.
Therefore, $\frac{1}{b_{n}}=\frac{1}{b_{1}}+(n-1) \cdot \frac{1}{4}=\frac{n}{4}$, hence $b_{n}=\frac{4}{n}\left(n \in \mathbf{N}^{*}\right)$.
Since $a_{n}=b_{n}-2$, we have $a_{n}=\frac{4}{n}-2=\frac{4-2 n}{n}\left(n \in \mathbf{N}^{*}\right)$.