$2x^2 + 2x + 3 \equiv 0 \pmod{p}$  has at most two solutions for primes $\geq 5$ Does 
have at most 2 solutions for primes $p\geq 5$? 
Mathematica confirms this for the first 10K primes: 
Max[ 
 Table[ 
  Length[ 
   Reduce[{ 
    Mod[2*x^2 + 2*x + 3, Prime[i]] == 0, 0 <= x <= Prime[i]}, x,Integers]], 
  {i,3,10000}] 
] 

Output: 2 

but I wasn't sure if it was true in general, and if it could be proved. 
Inspired by: Elementary proof that $2x^2+xy+3y^2$ represents infinitely many primes? 

You are trying to find the roots of a polynomial over the field of $p$-elements.
Over any integral domain, in particular over any field, there are at most two solutions.
In particular, there are at most two solutions (modulo $p$ of course; if $a$ is a solution, then so is $a+kp$ for any integer $k$). This holds for all primes, not not just those greater than or equal to $5$.
In fact, you can do better and say exactly for which primes there are two solutions. The usual quadratic formula works (when correctly interpreted) for all primes other than $p=2$. The discriminant of this quadratic is $-20 = 4(-5)$. So for $p\neq 2$, the quadratic has:


*

*No solutions if $-5$ is not a square modulo $p$;

*One solution if $p=5$ (since the discriminant is $0$ in that case);

*Two distinct solutions (modulo $p$) if $-5$ is a square modulo $p$.


When $p=2$, there are no solutions (the quadratic reduces to $3\equiv 0\pmod{2}$). When $p=5$, the quadratic reduces to $2x^2+2x-2\equiv 0 \pmod{5}$, or $2(x^2+x-1) =2(x^2-4x+4) = 2(x-2)^2\equiv 0\pmod{5}$. The unique solution is $x\equiv 2 \pmod{5}$.
For other primes, we need to determine if $-5$ is a square. This is easily done with quadratic reciprocity. For $p=3$, $-5 \equiv 1\pmod{5}$, so it is a square, and we have two distinct solutions (they are $x\equiv 0 \pmod{3}$ and $x\equiv  2\pmod{3}$). If $p\gt 5$, We have (using the Legendre symbol and Quadratic Reciprocity and its supplements):
$$\left(\frac{-5}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{5}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{p}{5}\right).$$
If $p\equiv 1\pmod{4}$, then $-1$ is a square modulo $p$, so $-5$ is a square if and only if $p$ is a square modulo $5$; that is, if $p\equiv 1,4\pmod{5}$. If $p\equiv 3\pmod{4}$, then $-1$ is not a square modulo $p$, so $-5$ is a square if and only if $p$ is not a square modulo $5$, that is, if $p\equiv 2,3\pmod{5}$.
So, there are two solutions modulo $p$ if and only if $p\equiv 1,3,7,9\pmod{20}$, and no solutions modulo $p$ if and only if $p\equiv 11,13,17,19\pmod{20}$.
In summary,

$2x^2 + 2x + 3 \equiv 0 \pmod{p}$, with $p$ a prime, has:
  
  
*
  
*No solutions if $p=2$, or if $p\equiv 11, 13, 17$, or $19\pmod{20}$.
  
*Exactly one solution modulo $p$ if $p=5$, namely $x\equiv 2 \pmod{5}$.
  
*Two distinct solutions modulo $p$ if $p\equiv 1 , 3, 7$ or $9\pmod{20}$; they are given by the quadratic formula modulo $p$. 
  

How do you use the quadratic formula? The usual way. The roots are given by
$$x = \frac{-2\pm\sqrt{-20}}{4} = \frac{-2\pm 2\sqrt{-5}}{4} = \frac{-1\pm\sqrt{-5}}{2}.$$
Given a prime $p$ for which $-5$ is a square, let $r$ be an integer such that $r^2\equiv -5\pmod{p}$. Let $s$ be an integer such that $2s\equiv 1\pmod{p}$ (say, $s=\frac{1-p}{2}$). Then the two solutions modulo $p$ are $x\equiv s(-1+r)\pmod{p}$ and $x=s(-1-r)\pmod{p}$. 
