$2x^2 + 3x +4$is not divisible by $5$ I tried by $x^2 \equiv 0, 1, 4 \pmod 5$ but how can I deal with $3x$? 
I feel this method does not work here.

$5$ is small. You could just consider all possibilities $\mod{5}$. There are only $0,1,2,3,4$. If any of these get you $0$, then, it can be possibly divisible by $5$. Otherwise, not. Let $f(x)=2x^2+3x+4$. So, $x\equiv0\implies f(0)\equiv4$, $x\equiv1\implies f(1)\equiv 2+3+4\equiv4\pmod{5}$, $x\equiv2\implies f(2)\equiv8+6+4\equiv3$, $x\equiv3\implies f(3)\equiv 18+9+4\equiv 1\pmod{5}$, and finally $x\equiv4\implies f(4)\equiv 32+12+4\equiv3$. An even faster way to do the calculations would just be to see that $3\equiv-2$ and $4\equiv-1$.
