Example 6 Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=0, a_{n+1}=5 a_{n}+\sqrt{24 a_{n}^{2}+1}(n \geqslant 1)$, try to find the general term formula of the sequence $\left\{a_{n}\right\}$.
Solving, we move terms and square (5), rearranging to get
\[a_{n+1}^{2}-10 a_{n} \cdot a_{n+1}+\left(a_{n}^{2}-1\right)=0\]
By swapping the indices in (6) and rearranging, we get
\[a_{n-1}^{2}-10 a_{n} \cdot a_{n-1}+\left(a_{n}^{2}-1\right)=0\]
From (6) and (7), we see that $a_{n+1}$ and $a_{n-1}$ are the roots of the equation
\[x^{2}-10 a_{n} x+\left(a_{n}^{2}-1\right)=0\]
Thus, we have
\[a_{n+1}=10 a_{n}-a_{n-1} \quad (n \geqslant 2)\]
The characteristic equation of (8) is $x^{2}-10 x+1=0$, with roots $x=5 \pm 2 \sqrt{6}$. Therefore, we can assume
\[a_{n}=\alpha_{1}(5+2 \sqrt{6})^{n}+\alpha_{2}(5-2 \sqrt{6})^{n}\]

Given $a_{1}=0, a_{2}=1$, we have
\[\alpha_{1}(5+2 \sqrt{6})+\alpha_{2}(5-2 \sqrt{6})=0\]
\[\alpha_{1}(5+2 \sqrt{6})^{2}+\alpha_{2}(5-2 \sqrt{6})^{2}=1\]

Solving, we get
\[\alpha_{1}=\frac{\sqrt{6}}{24}(5-2 \sqrt{6}), \quad \alpha_{2}=-\frac{\sqrt{6}}{24}(5+2 \sqrt{6})\]
Thus,
\[a_{n}=\frac{\sqrt{6}}{24}\left[(5+2 \sqrt{6})^{n-1}-(5-2 \sqrt{6})^{n-1}\right]\]