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 know $b^{2}-4ac=9$ with $a,b,c$ integers, then the equation $ax^{2}+bx+c (1)$ has rational roots.
Factorization of $ax^{2}+bx+c$ will be $a(x-t)(x-l)$ with $l,t \in \mathbb{Q}$. Now I can simplify to become every coefficient into integer number. It is the equation $(px+q)(rx+s)$.
$\bar{abc} = 100a+10b+c$, it is equal the equation (1) with $x=10$, then it is equal $(10p+q)(10r+s)$.
We conclude the number cannot be prime.
