$2x^2 + 3x +4$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.

Suppose $x\equiv 0\bmod 5$
Then $2x^2+3x+4\equiv4\not\equiv0\bmod 5$
Suppose $x\equiv 1\bmod 5$
Then $2x^2+3x+4\equiv 2+3+4\equiv4\not\equiv0\bmod 5$
Suppose $x\equiv 2\bmod 5$
Then $2x^2+3x+4\equiv 8+6+4\equiv 3\not\equiv0\bmod 5$
Suppose $x\equiv 3\bmod 5$
Then $2x^2+3x+4\equiv3+4+4\equiv1\not\equiv0\bmod 5$
Suppose $x\equiv 4\bmod 5$
Then $2x^2+3x+4\equiv 2+2+4\equiv 3\not\equiv0\bmod 5$
Then by exhaustion of the possible cases modulo $5$, we can conclude that $5\not |2x^2+3x+4$
