1. "For real roots, #b^2-4ac>0#:"
A basic quadratic has the formula #ax^2+bx+c#. When you solve by completing the square (which is just the rearranged quadratic formula) you end up square rooting #b^2-4ac#. (the discriminant).
If #Delta# (the discriminant) is less than zero, you are rooting a negative number (which is technically impossible) so you get a complex number and use #i#.
If the discriminant equals zero, then it disappears (root 0 = 0) and you only have one real root (there could be imaginary roots)
If #Delta# is bigger than zero, you square root a positive number, and end up with two real roots (because it's #±sqrt#).
2. How this relates to your question, and #k(k-4)>0#
In your equation, b=c; both are the same, k. So the discriminant is #k^2-(4*1*k) = k^2-4k = k(k-4)#, which must be bigger than zero to have two positive real roots as explained above.
3. the union stuff
This is basically a short way of writing what's stated on the 2nd and 3rd #=># symbols in your image.
#(-"infinity",0)#: k can be from negative infinity to zero. #(4,"infinity")#: k can also be anywhere bigger than 4.
#uu# means that these sets are disjoint, there's a gap between them. I think. Someone should double check this bit, I think it's right but I've never really learnt union notation.
