Example 4 Let real numbers $a, b$ satisfy
$$
3 a^{2}-10 a b+8 b^{2}+5 a-10 b=0 \text {. }
$$

Find the minimum value of $u=9 a^{2}+72 b+2$.
Notice,
$$
\begin{array}{l}
3 a^{2}-10 a b+8 b^{2}+5 a-10 b=0 \\
\Rightarrow(a-2 b)(3 a-4 b+5)=0 \\
\Rightarrow a-2 b=0 \text { or } 3 a-4 b+5=0 .
\end{array}
$$
(1) $a-2 b=0$.

Then $u=9 a^{2}+72 b+2=36 b^{2}+72 b+2$ $=36(b+1)^{2}-34$.
Thus, when $b=-1$, the minimum value of $u$ is -34.
$$
\text { (2) } 3 a-4 b+5=0 \text {. }
$$

Then $u=9 a^{2}+72 b+2=16 b^{2}+32 b+27$
$$
=16(b+1)^{2}+11 \text {. }
$$

Thus, when $b=-1$, the minimum value of $u$ is 11. In summary, the minimum value of $u$ is -34.