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
$$0&lt;\left(20a+b-d\right)&lt;\left(20a+b+d\right)\leq 20\cdot 9+9+9\leq 198.$$
Hence $a=1$, and 
$$\left(20a+b-d\right)&lt;\left(20a+b+d\right)\leq 20\cdot 1+9+9\leq 38.$$
Therefore $p&gt;100$ can not divide any of them and we have a  contradiction!
$c$ can take only $1, 3, 7, 9$ . [$abc$ is prime]
$b$ can take only $1, 2, 3, 4, 5, 6, 7, 8, 9$. [$b\neq 0$ as $-4ac$ can't be a square of a number]
$a$ can take $1, 2, 3, 4, 5, 6, 7, 8, 9$.
$b^2-k^2=4ac$.
Thus $b$ and $k$ both should be both odd or both even and $b&gt;k$ as $c$ can't be $&lt;0$.
$4ac=\text{Multiple of 4,12,28 or 36}$ as $c$ can take only $1, 3, 7, 9$.
$$(b,k)=(2,0),(4,0),(6,0),(8,0),(4,2),(6,2),(8,2),(6,4),(8,4),(8,6)(3,1),(5,1),(7,1),(9,1),(5,3),(7,3),(9,3),(7,5),(9,5),(9,7).$$
For all this values of $b$ and $k$, $a$ and $c$ can't assume such values such that $abc$ becomes a prime number. Thus, $b^2-4ac$ is never a perfect square when $abc$ is prime.