Example 4 Convert the pure periodic continued fraction
$$
2+\frac{1}{3}+\ldots=2+\frac{1}{3}+\frac{1}{2}+\frac{1}{3}+\cdots=[\overline{2,3}]
$$

into a quadratic irrational number.
Explanation: Let $x=[\overline{2,3}]$, then $x=2+\frac{1}{3}+\frac{1}{x}$. Therefore, we have Table 4.
Table 4
\begin{tabular}{|c|c|c|c|c|c|}
\hline$a_{n}$ & & & 2 & 3 & $x$ \\
\hline$p_{n}$ & 0 & 1 & 2 & 7 & $7 x+2$ \\
\hline$q_{n}$ & 1 & 0 & 1 & 3 & $3 x+1$ \\
\hline
\end{tabular}

Thus, $x=\frac{7 x+2}{3 x+1}$, which means $3 x^{2}-6 x-2=0$. Solving this, we get $x=\frac{3+\sqrt{15}}{3}$.