Given that the $3$-digit number $abc= 100a+10b+c$ (not to be confused with $a*b*c$) is a prime, $p$, prove that $b^2-4*a*c$ cannot be a perfect square.
So far, I tried to prove it by contradiction. Let $f(x)=ax^2+bx+c$, so that $f(10)=p$, and both roots of $f(x)$ must be negative integers (since $a,b,c$ are digits). Then, I substituted the values in for $p$, and got $p=a(10-x_1)(10-x_2)$, where $x_1, x_2$ are the roots of $f(x)$. $p$ must divide one of the factors, say $(10-x_1)$, but I cannot continue from this point.
If $b^2 - 4ac$ is a perfect square, considering the quadratic formula or its derivation, we see that the quadratic equation you mentioned, $ax^2 + bx + c$, has two negative rational roots. Hence we can write
$$f(x) = ax^2 + bx + c = (kx + l)(mx +n )$$
for positive integers $k,l,m,n$. We have $f(10) = p = (10k+l)(10m+n)$, so $p$ must have two 2-digit factors. This contradicts the fact that $p$ is prime.