Showing that there are no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=9$. 
Problem: Show that there are no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=9$.
  Solution. Upon the given condition, the quadratic equation $ax^2+bx+c$ can be written as $(px+q)(rx+s)$ and we have:
   $$\overline{abc}=100a+10b+c=(10p+q)(10r+s)$$
  i.e. $\overline{abc}$ is not a prime which is a contradiction, so there is no such prime available.  

I don't get the reasoning completely.

We show a more general fact:

There is no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=d^2$ where $d$ is a non-negative integer.

We first note that $d\leq b\leq 9$. Since $b^2-4ac=d^2$, we have that
$$P(x):=ax^2+bx+c=a\left(x-\frac{-b+d}{2a}\right)\left(x-\frac{-b-d}{2a}\right)$$
Assume that $P(10)=\overline{abc}=p$ where $p$ is some prime of three digits. Then 
$$4ap=\left(20a+b-d\right)\left(20a +b+d\right)$$
which implies that $p$ divides $\left(20a+b-d\right)$ or $\left(20a+b+d\right)$. 
Now $1\leq a\leq 9$, $0\leq b\leq 9$, and
$$0<\left(20a+b-d\right)<\left(20a+b+d\right)\leq 20\cdot 9+9+9\leq 198.$$
Hence $a=1$, and 
$$\left(20a+b-d\right)<\left(20a+b+d\right)\leq 20\cdot 1+9+9\leq 38.$$
Therefore $p>100$ can not divide any of them and we have a  contradiction!
