Answer: 10

Small Hint:

Let the two integer roots be $r$ and $s$, and eliminate $a$ using Vieta’s formulas

Big Hint:

Complete a product after obtaining $rs=-6(r+s)$

Solution:

Let the integer roots be $r$ and $s.$ Vieta’s formulas give $r+s=-a$ and $rs=6a,$ so $rs=-6(r+s).$ Therefore $$(r+6)(s+6)=36.$$ Conversely, every ordered integer factorization $uv=36$ gives integer roots $r=u-6,$ $s=v-6,$ and $a=12-u-v.$ Unordered positive factor pairs of $36$ have sums $37,$ $20,$ $15,$ $13,$ $12,$ and the corresponding negative factor pairs have their negatives as sums. These ten sums are distinct, so they give $10$ distinct values of $a.$