$2x^2 + 3x +4$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.
 A: When you want to get rid of an $x$ term in a quadratic, you should always try completing the square! We find that $$2x^2+3x+4=2((x+2)^2-2)$$ If this is $0\pmod{5}$, then so is $(x+2)^2-2$. But you've already pointed out that $\not\exists k\in\mathbb Z$ such that $k^2=2\pmod{5}$ so that's impossible.
A: The quadratic formula works in all fields whose characteristic is not $2$. Since you are working in $\mathbb{Z}_5$ we are safe.
The quadratic formula gives that the solutions would be:
$$ x = \frac{-3 \pm \sqrt{3^2 - 4(2)(4)}}{8} = \frac{-3 \pm \sqrt{-23}}{8} = \frac{-3 \pm \sqrt{2}}{8}_.$$
You can manually check that $\sqrt{2}$ does not exist in $\mathbb{Z}_5$. So no roots exist in $\mathbb{Z}_5$.
A: $5$ 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 $4\equiv-1$.
A: The method does work. How did you deal with $2x^2$ anyway? You know that $x^2 \equiv 0, 1, 4 \pmod 5$. So you just double those, like this: $0, 2, 8$, rewrite as $0, 2, 3$.
Likewise with $3x$, you just have to triple $0, 1, 2, 3, 4$ to get $0, 3, 6, 9, 12$ which we rewrite as $0, 3, 1, 4, 2$.
And lastly $4 \equiv 4 \pmod 5$, obviously.
So the possibilities are:


*

*If $x \equiv 0 \pmod 5$, then $2x^2 + 3x + 4 \equiv 0 + 0 + 4 = 4$.

*If $x \equiv 1 \pmod 5$, then $2x^2 + 3x + 4 \equiv 2 + 3 + 4 = 9 \equiv 4$.

*If $x \equiv 2 \pmod 5$, then $2x^2 + 3x + 4 \equiv 8 + 6 + 4 = 18 \equiv 3$.

*If $x \equiv 3 \pmod 5$, then $2x^2 + 3x + 4 \equiv 18 + 9 + 4 = 31 \equiv 1$.

*If $x \equiv 4 \pmod 5$, then $2x^2 + 3x + 4 \equiv 32 + 12 + 4 = 48 \equiv 3$.


As we failed to find a $0$ this way, we conclude that $2x^2 + 3x + 4$ is never divisible by $5$. Not as elegant as completing the square, but easy enough for a child to do.
A: 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 $5\not |2x^2+3x+4$
A: $$5|\;(2x^2+3x+4)\iff$$ $$ \iff5|\;3(2x^2+3x+4)=(6x^2+9x+12)\iff$$ $$\iff 5|\;((6x^2+9x+12)-(5x^2+5x+10))=$$ $$=(x^2+4x+2)=(x+2)^2-2.$$ But no square is $2$ more than a multiple of $5.$
