8. How many real numbers $a$ are there such that $x^{2}+a x+6 a$ $=0$ has only integer solutions?

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8. Let $x^{2}+a x+6 a=$ O have integer solutions $m, n$ $(m \leqslant n)$. Then we have
$$
a=-(m+n), \quad 6 a=m n .
$$

Since $a$ must be an integer, we also have
$$
-6(m+n)=m n \text { . }
$$

H $(m+6)(n+6)=36$. Since $36=36 \times 1=18 \times 2=12 \times 3=6 \times 6$, the solutions 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 $49,32,27,25$,
$$
{ }^{2} 4,-25,-8,-3,-1,0 \text {. A total of 10. }
$$