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

$$5|\;(2x^2+3x+4)\iff$$ $$ \iff5|\;3(2x^2+3x+4)=(6x^2+9x+12)\iff$$ $$\iff 5|\;((6x^2+9x+12)-(5x^2+5x+10))=$$ $$=(x^2+4x+2)=(x+2)^2-2.$$ But no square is $2$ more than a multiple of $5.$
