Example 5 Solve the equation
$$
5 x^{2}+10 y^{2}-12 x y-6 x-4 y+13=0 \text {. }
$$
(1994, Tianjin City Junior High School Mathematics Competition)
Solution: Rearrange the original equation as an equation in terms of $x$, we get
$$
5 x^{2}+(-12 y-6) x+\left(10 y^{2}-4 y+13\right)=0.
$$

For the original equation to have real solutions, it is necessary that
$$
\begin{aligned}
\Delta & =(-12 y-6)^{2}-4 \times 5\left(10 y^{2}-4 y+13\right) \\
& =-56(y-2)^{2} \geqslant 0,
\end{aligned}
$$

Solving this, we get $y=2$. Substituting $y=2$ back into the original equation and simplifying, we get $(x-3)^{2}=0$, which means $x=3$.
$\therefore$ The real solutions to the original equation are $x=3, y=2$.