Example 11 (9th American Invitational Mathematics Examination) How many real numbers $a$ are there such that $x^{2}+a x+6 a=0$ has only integer solutions.

保留了原文的换行和格式。
Let the equation $x^{2}+a x+6 a=0$ have integer solutions $m, n(m \leqslant n)$, then we have
$$
\begin{array}{l}
m+n=-a, \\
m n=6 a .
\end{array}
$$

Thus, $-6(m+n)=m n$, which means
$$
(m+6)(n+6)=36 \text {. }
$$

Since $36=1 \cdot 36=2 \cdot 18=3 \cdot 12=4 \cdot 9=6 \cdot 6$, the solutions $(m, n)$ satisfying $m \leqslant n$ are
$$
\begin{array}{l}
(-42,-7),(-24,-8),(-18,-9),(-15,-10), \\
(-12,-12),(-5,30),(-4,12),(-3,6),(-2,3),(0,0) .
\end{array}
$$

The corresponding values of $a=-(m+n)$ are
$$
49,32,27,25,24,-25,-8,-3,-1,0 \text {. A total of } 10 \text { values. }
$$