I assume the question is "find conditions that are necessary and sufficient to guarantee solutions" rather than "find necessary conditions and also find sufficient sufficient conditions for a solution." If the former is the case, then you're asked for constraints on $a$ and $b$ such that (1) if the conditions are met then $ax^2+b$ is zero for some $x$ and (2) if the conditions are not met then $ax^2+b$ isn't zero for any $x$.
So, when will $ax^2+b$ have some value $x$ for which the expression is zero? Well, as André suggested, try to solve $ax^2+b=0$ "mechanically". By subtracting we have the equivalent equation $ax^2 = -b$ and we'd like to divide by $a$ to get $x^2=-b/a$. Unfortunately, we can't do that if $a=0$, so we need to consider two cases
- $a=0$
- $a\ne 0$
In the first case, our equation becomes $0\cdot x^2+b=0$, namely $b=0$ and if $b=0$ (and, of course $a=0$) then any $x$ will satisfy this. In other words, there's a solution (actually infinitely many solutions) if $a=b=0$, as you've already noted.
Now, in case (2) we can safely divide by $a$ to get $x^2=-b/a$. When does this have a solution? We know that $x^2\ge 0$ no matter what $x$ is, so when will our new equation have a solution? You said you don't want the full answer, so I'll denote the answer you discover by $C$, which will be of the form "some condition on $a\text{ and }b$". Once you've done that, your full answer will be:
$ax^2+b=0$ has a solution if and only if
- $a=b=0$, or
- Your condition $C$.
These are sufficient, since either guarantees a solution, and necessary, since if neither is true, then there won't be a solution (since we exhausted all possibilities for solutions).