For this equation to be solvable, both $f(x)\equiv 0\pmod{3}$ and $f(x)\equiv 0\pmod{5}$ need to be solvable.

Considering $\pmod{3}$ first. $$f(x)=4x^2 + 27x-9\equiv x^2 \equiv 0\pmod{3} \implies x\equiv 0\pmod{3}$$

This means that $f(x)\equiv 0\pmod{3}$ has only one solution $x\equiv 0\pmod{3}$. Now setting $x=0$, an element in this residue class, to $f(x)\equiv 0\pmod{5}$ yields: $$f(0)=-9\equiv 1\not\equiv 0\pmod{5}$$

This implies that $f(x)\equiv 0\pmod{3}$ and $f(x)\equiv 0\pmod{5}$ cannot hold simultaneously. Hence, we conclude that $f(x)\equiv 0\pmod{15}$ is insolvable.