Polynomial Division

We are finding the number of values of $x$ such that $5x^2+4x+10\equiv0\text{ mod }4.$ $\begin{aligned} 5x^2+4x+10&\equiv0\text{ mod }4\\ x^2+2&\equiv0\text{ mod }4\\ x^2&\equiv-2\text{ mod }4\\ x^2&\equiv2\text{ mod }4 \end{aligned}$ However, this leads to a contradiction because the quadratic residues mod $4$ are $0$ and $1.$ Thus, there are no integers such that $5x^2+4x+10\equiv0\text{ mod }4.$

We can create cases.

$\text{Case 1:}$

$x$ is an even number. Then we can write $x$ of the form $2a$ . Then the polynomial changes to $5(2a)^2+4(2a)+10=20a^2+8a+10$ .

Now we can see that $20a^2$ is divisible by $4$ and $8a$ is also divisible by $4$ , but $10$ is not divisible by $4$ . This means the polynomial is not divisible by $4$ for any even number.

$\text{Case 2:}$

$x$ is an odd number. Then we can write $x$ of the form $2a+1$ . Then the polynomial changes to $5(2a+1)^2+4(2a+1)+10=5(4a^2+4a+1)+4(2a+1)+10\\=20a^2+20a+5+8a+4+10=20a^2+28a+19$ .

Here we can see that $20a^2$ is divisible by $4$ and $28a$ is also divisible by $4$ , but $19$ is not. Hence, no odd number will satisfy this.

Thus there are no solutions.