Example 2 Let the parabola
$$
y=x^{2}+(2 a+1) x+2 a+\frac{5}{4}
$$

intersect the $x$-axis at only one point.
(1) Find the value of $a$;
(2) Find the value of $a^{18}+323 a^{-6}$.
(1998, National Junior High School Mathematics Competition)
Solution: (1) From the given condition, the equation
$$
x^{2}+(2 a+1) x+2 a+\frac{5}{4}=0
$$

has two equal real roots, hence
$$
\Delta=(2 a+1)^{2}-4\left(2 a+\frac{5}{4}\right)=0,
$$

which simplifies to $a^{2}-a-1=0$.
Solving for $a$, we get $a=\frac{1 \pm \sqrt{5}}{2}$.
(2) From (1), we know $a^{2}-a-1=0$, thus
$$
\begin{array}{l}
a-a^{-1}=1, \\
a^{2}+a^{-2}=3, \\
\begin{aligned}
a^{6}+a^{-6} & =\left(a^{2}+a^{-2}\right)^{3}-3\left(a^{2}+a^{-2}\right) \\
& =3^{3}-3 \times 3=18, \\
a^{12}+a^{-12} & =\left(a^{6}+a^{-6}\right)^{2}-2 \\
& =18^{2}-2=322 .
\end{aligned}
\end{array}
$$

Substituting $a^{12}+a^{-12}$ with 322, we get
$$
\begin{aligned}
323 & =a^{12}+a^{-12}+1 \\
\therefore & a^{18}+323 a^{-6} \\
= & a^{18}+\left(a^{12}+a^{-12}+1\right) a^{-6} \\
= & \left(a^{18}+a^{-18}\right)+\left(a^{6}+a^{-6}\right) \\
= & \left(a^{6}+a^{-6}\right)\left(a^{12}-1+a^{-12}\right) \\
& +\left(a^{6}+a^{-6}\right) \\
= & \left(a^{6}+a^{-6}\right)\left(a^{12}+a^{-12}\right) \\
= & 18 \times 322=5796 .
\end{aligned}
$$