jmc

Let $r$ and $s$ be the integer roots. Then by Vieta's formulas, $r+s=-a$ and $rs=6a.$ Thus, $$rs+6(r+s)=0.$$Applying Simon's Favorite Factoring Trick, we have $$rs+6(r+s)+36=36⟹(r+6)(s+6)=36.$$The number $36=2^2{3}^2$ has $2(2+1)(2+1)=18$ factors, both positive and negative; they come in $8$ pairs, with singletons $6$ and $-6.$ However, since the order of $r$ and $s$ does not matter, each pair should only be counted once, so there are $8+1+1=\boxed{10}$ possible values for $a.$