I have a quadratic polynomial $p(x)=a-bx+cx^2$, where $a,b,c$ are positive integers. If I assume that $a$ is odd, $b$ is even, and that the discriminant is negative (so that the polynomial is irreducible), is it possible to argue that $p(x)$ is a prime for at least some $x$? Or are there polynomials of this type that are always composite?
Note: in my case, $p(x)$ is odd for all $x$. Does this change anything?
Since you technically didn't say it had to be irreducible, something like
$$3-6x+9x^2$$
works. Its discriminant is $6^2-4\cdot3\cdot9=-72$ is negative, $a$ is odd, and $b$ is even. However, every value is a multiple of $3$.
If there exists no prime that divides each coefficient, I will show that there exists no prime $p$ that divides every value of the polynomial (this does in no way show that there must always exist a prime, but it likely precludes most elementary attempts to find a counterexample):
If
$$ax^2-bx+c\equiv 0\bmod p$$
for all positive integer $x$, they are equivalent as polynomials. If $p\geq 3$, these two are distinct polynomials, and if $p=2$, we require that $2|c$ (at $x=2$), which is one of the conditions you required cannot occur.
As darij grinberg noted in his comment, if one could prove your conjecture one could show that there are infinitely many positive integers $n$ for which $n^2+1$ is prime. Indeed, assume there are only finitely many, and let them be $n_1,\cdots,n_m$. Let $k$ be a positive integer that does not divide $n_i+2$ for any $1\leq i\leq m$. Then
$$(kx-2)^2+1=k^2x^2-2kx+5$$
is composite for all positive integer $x$ and satisfies all your criteria.
In other words, proving that your conjecture is true is quite hard.