Example 11. Solve the equation
$$
5 x^{2}+10 y^{2}-72 x y-6 x-4 y+13=0 \text {. }
$$
(1994, Tianjin Junior High School Mathematics Competition)
Solve: The original equation can be transformed into a quadratic equation in terms of $x$
$$
5 x^{2}+(-12 y-6) x+10 y^{2}-4 y+13=0 \text {. }
$$

Since $x$ is a real number, we have
$$
\begin{aligned}
\Delta & =(-12 y-6)^{2}-4 \times 5\left(10 y^{2}-4 y+13\right) \\
& =-(y-2)^{2} \geqslant 0 .
\end{aligned}
$$

Solving this, we get $y=2$.
Substituting $y=2$ into the original equation, we get
$$
x^{2}-6 x+9=0 \text {. }
$$

Solving this, we get $x=3$.
Therefore, the real solutions to the original equation are $x=3$ and $y=2$.