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.
 A: 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!
A: 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.
A: Try some of these with the $p,q,r,s$ letters, see what happens. I suggest $299$ and $451$ as there will be just one way to get a product.
Here is one $252.$ $ \; \; 2x^2 + 5x + 2 = (2x+1)(x+2).$ $ \; \; \;252 = 21 \cdot 12$
130 = 2 * 5 * 13
154 = 2 * 7 * 11
230 = 2 * 5 * 23
252 = 2^2 * 3^2 * 7
275 = 5^2 * 11
299 = 13 * 23
330 = 2 * 3 * 5 * 11
396 = 2^2 * 3^2 * 11
430 = 2 * 5 * 43
451 = 11 * 41
530 = 2 * 5 * 53
572 = 2^2 * 11 * 13
630 = 2 * 3^2 * 5 * 7
693 = 3^2 * 7 * 11
730 = 2 * 5 * 73
830 = 2 * 5 * 83
930 = 2 * 3 * 5 * 31
992 = 2^5 * 31

A: NOTE: I fixed the problem that originally occured, so I have removed the disclaimer again. But the proof got a bit more complicated in order to establish that $a$ must be a multiple of $nv$.

Let me provide the proof of the following result, which should connect the dots in the original solution given in the OP:

If $a,b,c$ are integers and $ax^2+bx+c$ has rational roots, then
  $$
ax^2+bx+c=(px+q)(rx+s)
$$

for some integers $p,q,r$, and $s$. To see this, assume the two roots to be $\tfrac mn$ and $\tfrac uv$ where both fractions have been simplified to their lowest terms. Then we have
$$
\begin{align}
ax^2+bx+c&=a(x-\tfrac mn)(x-\tfrac uv)\\
&=ax^2-a(\tfrac mn+\tfrac uv)x+a\tfrac mn\tfrac uv
\end{align}
$$
Hence it follows that
$$
\begin{align}
b&=-a\tfrac{mv+nu}{nv}\\
c&=a\tfrac{mu}{nv}
\end{align}
$$
Now suppose both fractions above could be expressed in terms of some common denominator $nv/p$ for some prime $p$. Combining the facts that $m/n$ and $u/v$ are in their lowest terms, but $\frac{mu}{nv}$ is not, it then follows by Euclid's Lemma, that $p$ divides either the pair $n,u$ or the pair $m,v$. In either case it cannot divide the other numerator $mv+nu$ since it divedes one of the terms but is relatively prime to the other. Hence no such common factor $p$ exists. Thus $nv$ is the least common denominator, and therefore $a$ must be a multiple of $nv$.
Now write $a=\lambda nv$. Then it follows that
$$
\begin{align}
a(x-\tfrac mn)(x-\tfrac uv)
&=\lambda nv(x-\tfrac mn)(x-\tfrac uv)\\
&=(\lambda nx-\lambda m)(vx-u)
\end{align}
$$
which has the form $(px+q)(rx+s)$ with $p,q,r$, and $s$ all integers.

By the way, this easily provides the foundation for reaching the general result:

Show that $\overline{abc}$ with $b^2-4ac=d^2$ where $abc=100a+10b+c$ is a $3$-digit number and $d$ is a natural number can never be a prime number.

Solution: Since $d^2$ is a perfect square, the roots of $ax^2+bx+c$ must be rational. They must also be non-positive, since $d^2\leq b^2$, and thus we have
$$
ax^2+bx+c=(px+q)(rx+s)
$$
for some natural numbers $p,q,r$, and $s$. Plugging in $x=10$ we get
$$
100a+10b+c=(10p+q)(10r+s)
$$
proving that $\overline{abc}$ can be non-trivially factored into two factors greater than or equal to $10$.
A: $c$ 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 $b>k$ as $c$ can't be $<0$.
$4ac=\text{Multiple of 4,12,28 or 36}$ 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.
