Boring part
For the equation ax^2 + bx + c = 0, the discriminant is b^2-4ac, and if this number is positive then there are two real roots. So you can figure out what a, b and c are in this particular equation (don't let the ks put you off, the process is the same), then show that this must be positive for real k.
Interesting part
Why is this true?
Because when you complete the square, you get
x^2 + bx/a + c/a = 0
(x + b/2a)^2 - b^2/4a^2 + c/a = 0
(x + b/2a)^2 = b^2/4a^2 - c/a
= b^2/4a^2 - 4ac/4a^2
= (b^2 - 4ac)/(4a^2)
So
(x + b/2a)^2 = (b^2 - 4ac)/(4a^2)
4a^2*(x + b/2a)^2 = (b^2 - 4ac)
[2a*(x + b/2a)]^2 = b^2 - 4ac
2a*(x + b/2a) = +/- sqrt(b^2 - 4ac)
That last line tells you why it is true. If b^2-4ac is positive then its square root, the RHS of this equation, is real and we have real solutions for x. If it is negative then the RHS is an imaginary number, and we have no real solutions. So b^2 - 4ac is the key.