The 'discriminant' is the b^2 - 4ac portion of the Quadratic Formula:
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
a = k-5 b = -k c = 5
When the discriminat = 0 there is only one solution to the quadratic
(-k)^2 - 4 (k-5)(5) = 0 <======== can you take it from here....solve for k ?