14. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{3}{2}, a_{n+1}=\frac{a_{n}^{2}+4}{2 a_{n}+3}$, find its general term formula $a_{n}$.
14. Solution: It is easy to know that the fixed point equation of the sequence is $x=\frac{x^{2}+4}{2 x+3}$, i.e., $x^{2}+3 x-4=0$, solving which gives $x_{1}=-4$, $x_{2}=1$.
Therefore, $\frac{a_{n+1}+4}{a_{n+1}-1}=\frac{\frac{a_{n}^{2}+4}{2 a_{n}+3}+4}{\frac{a_{n}^{2}+4}{2 a_{n}+3}-1}=\left(\frac{a_{n}+4}{a_{n}-1}\right)^{2}$.
Let $b_{n}=\frac{a_{n}+4}{a_{n}-1}$, then $b_{1}=11$, and $b_{n}=b_{n-1}^{2}=b_{n-2}^{2^{2}}=b_{n-3}^{2^{3}}=\cdots=b_{1}^{2^{n-1}}=11^{2^{n-1}}$.
Thus, $a_{n}=\frac{11^{2^{n-1}}+4}{11^{2^{n-1}}-1}\left(n \in \mathbf{N}^{*}\right)$.