Solving quadratic modulo congruence: review. I've been trying to solve a quadratic modulo congruence and I think I have the right solution, but in the end there's two things:
1) I cannot explain how $\sqrt(12)$ divided by 2 mod 23 is equal to $\sqrt(3)$. By pure chance I looked at another link to try and understand quadratic modulo function and there's where I took this from (http://web.science.mq.edu.au/~chris/numbers/CHAP03%20Quadratic%20Congruences.pdf)
2) In the end I have two solutions I got to the second one by some trial and error. So I would like to improve this for future scenarios.
Let me explain my rationale:
The congruence is: $$ {x}^2 +4{x} + 1 \equiv 0 \pmod {23}$$
I get this into a quadratic formula
$${x} = \frac {-4 \pm \sqrt{4^2 -4*1*1}}2 $$
which simplifies to:
$${x} = \frac {-4 \pm \sqrt{12}}2 $$
I know $\sqrt{12}$ is valid in$\pmod {23}$ since $gcd(12,23)=1$. However the calculations of the square roots made me think the proble was not being solved correctly, and then on the link provided above I saw a simplification of ${\sqrt{12}\over2} = \sqrt{3}$, which made me try again
Question: I know that dividing by 2 is the same as the multiplicative inverse. So I played with the idea of $\sqrt{12} * 2^{-1} = 12^{-1} * 2^{-1}$ but still I didn't see how this could become $\sqrt{3}$
Assuming this is correct, I proceeded by simplifying the congruence to:
$$x = -2 \pm \sqrt{3}$$
Manually I found that 7 was a square root of 3, and so I proceeded with:
$$x = -2 + 7 \land x= -2 -7 $$ 
$$x = 5 \land x= -9 $$ 
I tested both solutions and both work. I didn't use -9, I used -14, since I used the positive MOD of -7.
$$x = 5 \land x= 14 $$  
So this worked, both solutions were valid testing against the original formula, but I'm left with quite a few questions:


*

*The mentioned above division of the square root?

*Should I have tried all manual square roots of 3 until 23.

*Do I have more solutions?


And to finish, if there's anything left I didn't mention and can be improved in my process, please let me know.
I hope this is a valid question.
Kind regards.
Update: I've updated the question due to insightful comments that addressed some of the questions.
 A: I'll try to help. It appears you have some concern about root 12 in mod 23. I do not know if that concern is justified in general. However ,
$ \sqrt{12} = 9 \ \ or \ 14 $
$ \sqrt{4} = 2 \ or \ 21$
$\sqrt{3} = 7 \ or \ 16 $
$ \sqrt{4} \cdot \sqrt{3} = (2 \cdot 7) \ or \ (2 \cdot 16) \ or \ (21 \cdot 7) \ or \ (21 \cdot 16) $
$\sqrt{4} \cdot \sqrt{3} = 14 \ or \ 9 \ or \ 9 \ or \ 14 $
So it looks like we can split up root 12 ,
$ \frac{-4 \ \pm \ \sqrt{12} }{2} = -2 \ \pm \ \frac{\sqrt{4} \cdot \sqrt{3} }{ \sqrt{4}} = -2 \ \pm \ \sqrt{3} $
You can bypass the quadratic formula. Complete the square by adding 3 to both sides of the congruence.
You can always complete the square on any quadratic congruence.
UPDATE: For question 3 , there can be exactly 2 solutions since the mod is prime (23) and co-prime to the coefficient of x^2 (1).
I found a way to make your idea using $ 2^{-1} $ work , the inverse of 2 is 12 in (mod 23) because 
$ 2 \cdot 12 \equiv 1 \ (mod \ 23) $
We can then write 
$ (-4 \ \pm \ \sqrt{12} ) \cdot 2^{-1} = (-4 \ \pm \ \sqrt{12} ) \cdot 12 $
$ -48 \ \pm \ 12 \cdot \sqrt{12} = -2 \ \pm \ 12(9 \ or \ 14) $
Using 9 for root 12 gives the two solutions for x as you can check.
A: If you divide x^2 + 4x + 1 by 23, the only values of x that work are 5 and 14.
So x = 5 mod 23 and x = 14 mod 23.
