Range of values of $k$ for which the equation has distinct real roots First of all, let me state the problem and my rationale:
Show the range of values of $k$ for which $x^2-kx+2k+1=0$ has distinct real roots
Premise I: The discriminant is: $b^2 - 4ac$*
Premise II: $ k > 0$, otherwise if $k=0$ it has only one root and if $ k < 0$ then it's not in the real domain.
Then:


*

*$b^2 - 4ac > 0$

*$(-k)^2 -4(1 * 2k +1)$

*$k^2 - 8k - 4 > 0 $

*$(k-4)^2 - 4^2 - 4 > 0$ Completing the squares

*$(k-4)^2 - 20 > 0$ 

*$(k-4)^2 > 20$
At this point I can tell that this is zero if $ (k - 4)^2 = 20 $
Hence:


*$(k-4) > \pm\sqrt{20}$ 

*$k > 4 \pm2\sqrt{5}$
Therefore, the range is: 
$\{k: k < 4- 2\sqrt{5}\} \cup \{k: k > 4 +2\sqrt{5}\} $ 
because that's when $k^2 -8k-4 > 0$
Great. 
Now to the question:
What's the intuition behind my step 7? 
I learned it this way, but I am not sure about the "$\pm$" sign. Is it because I am transforming it again into the discriminant form or is it simply because $(-\sqrt{20})^2$ is a way to get $(k-4)$ back to $(k - 4)^2$. I'm not using any other intuition other than that here.
Thanks a lot in advance!
Edit-1: fixed question 2's typo.

Edit-2: Removing question 2. I found the solution for 1 and will create another post for that

 A: Step 7 is incorrect, because (as you showed) it leads to 
$k > 4 \pm2\sqrt{5}$,
which leads to 
$$k > 4- 2\sqrt{5} \quad\text{or}\quad k > 4 +2\sqrt{5},$$
from which you would infer that the range is
$\{k: k \color{red}{>} 4- 2\sqrt{5}\} \cup \{k: k > 4 +2\sqrt{5}\},$
which is not the correct range; the first $>$ should be $<$ instead.
The only way you got the correct range at the end was by ignoring what your formula actually said and writing $<$ instead of $>.$
When $(k-4)^2 = 20$ it is indeed OK to write $k - 4 = \pm\sqrt{20}$;
this gives you the two values of $k$ at which $k^2−8k−4$ is zero.
But you could equally well write $\lvert k - 4\rvert = \sqrt{20}$ with the same result.
The reason you can get away with either method for solving an equation is that one equation is equivalent to a second equation if the second equation merely reverses the signs of both sides of the first equation.
But reversing the signs of both sides of an inequation does not give you an equivalent inequation. You have to reverse the direction of the inequality as well.
One thing you could do instead of step 7, of course, is to use the solution of
the equation $(k-4)^2 = 20$ to factorize the polynomial $k^2−8k−4.$
You already have one answer demonstrating this.
Another approach (which is somewhat in the spirit of your step 7, but actually correct)
is to take $(k - 4)^2 > 20$ and deduce from this that 
$\lvert k - 4\rvert > \sqrt{20}$
(using the fact that the positive square roots of two equal positive numbers are equal).
The absolute value tells you that either
$$k - 4 > \sqrt{20} \quad\text{or}\quad -(k - 4) > \sqrt{20},$$
from which you can deduce that
$$k > 4 + 2\sqrt{5} \quad\text{or}\quad k < 4 - 2\sqrt{5},$$
which leads to the correct answer.

Update: A third approach is that you use the equation $(k-4)^2 = 20$
to determine that $k^2−8k−4 = 0$ when $k = 4 \pm 2\sqrt5,$
from which you conclude that $k^2−8k−4$ changes signs at 
$k = 4 - 2\sqrt5$ and at $k = 4 + 2\sqrt5.$
That is, $k^2−8k−4$ has the same sign wherever we put $k$ in the interval 
$(-\infty,4 - 2\sqrt5),$ 
it has the opposite sign when $k$ is in $(4 - 2\sqrt5,4 + 2\sqrt5),$
and it has the same sign as the first one in
$(4 + 2\sqrt5, \infty).$
By testing just one value of $k$ (for example $k=4$) you can determine that the signs are $+,-,+$ and therefore the intervals where $k^2−8k−4$ is positive
are $(-\infty,4 - 2\sqrt5)\cup(4 + 2\sqrt5, \infty).$
A: Step 7 is wrong as we cannot write that way rather it should be again like parabola and written in factor form as $$ {\ ( k-4-2\sqrt{5})(k-4+2\sqrt{5}) }> 0 $$
Hence either $${\ k < 4-2 {\sqrt{5}} } $$  or 
$$  {\ k > 4 + 2{\sqrt {5}} }  $$
