It's possible for positive a too?

Since $f(x) = ax^2+(3-2a)x - 6 >0$ , $f(x)$ has a continuous range of $x$ , where it is positive. The quadratic curve must be inverted and $a<0$ . Then, we have:

$\begin{aligned} f(x) & = ax^2+(3-2a)x - 6 \\ \implies x & = \frac {2a-3\pm \sqrt{4a^2-12a+9+24a}} {2a} \\ & = \frac {2a-3\pm \sqrt{4a^2 + 12a+9}} {2a} \\ & = \frac {2a-3\pm (2a + 3)} {2a} \\ & = \begin{cases} 2 \\ - \dfrac{3}{a} \end{cases} \end{aligned}$

There are two roots for $f(x)$ , 2 and $-\dfrac{3}{a}$ . Since $a< 0$ , $-\dfrac 3a > 0$ , this means that $-\dfrac 3a > 2$ . Because if $-\dfrac 3a < 2$ , then the three integral values of $x$ are -1, 0 and 1, which is not acceptable because $-\dfrac 3a > 0$ . It does not include $x=2$ because $f(2) = 0 \not > 0$ . Therefore, the three integral values of $x$ are 3, 4 and 5. Therefore,

$\implies 5 < -\dfrac{3}{a} \le 6 \implies -\dfrac{3}{5} < a \le -\dfrac{1}{2} \implies a \in \boxed{\left( -\dfrac{3}{5}, -\dfrac{1}{2} \right]}$