For how many 3-digit prime numbers $\overline{abc}$ do we have: $b^2-4ac=9$? 
For how many 3-digit prime numbers $\overline{abc}$ do we have: $b^2-4ac=9$?

The only analysis I did is:
$(b-3)(b+3)=4ac \implies\ b\geq3 $
$b=3\implies\ c=0\implies impossible!!$
So I deduced that $b\gt3$
Is there any better way for quickly solving this??
Or maybe a trick applies...

$(b-3)(b+3) = 4ac$
Now, $4ac$ is an even number. Hence, $(b-3)(b+3)$ is also and even number. That means, b cannot be even number else, $(b-3)(b+3)$ will if odd. 
Hence, $b > 3$ and b is odd. Thus, b can be 5, 7, 9. 
Putting, the values for b:
When $b = 5$, we get $4ac = (5-3)(5+3) = 2 * 8 = 16$ or $ac = 4 = 2*2 = 1*4$. No prime numbers possible with these combinations.
When $b = 7$, we get $4ac = (7-3)(7+3) = 4.10 = 40$ or $ac = 10 = 2*5$. Again, no combinations possible.
When $b = 9$, we get $4ac = (9-3)(9+3) = 6.12 = 72$ or $ac = 18 = 2*9 = 3*6$. Again, no combinations possible.
Thus, no such 3 digit prime number exists.
