The key to answering this question lies in finding the discriminant b^2-4ac. If b^2-4ac is greater than zero, the quadratic equation will have 2 real solutions.
Typing out y=x^2+kx-k, we see that a=1, b=k and c = -k. The discriminant in this case is
b^2 - 4(a)(c), or k^2 - 4(1)(-k), or k^2 + 4k
For which k is this k^2 + 4k = 0? Factor k^2 + 4k, obtaining k(k+4) = 0; the roots of this are k=0 and k = -4.
Now set up a number line for k, drawing empty circles at k = 0 and k = -4. These two values divide the number line into three intervals:
(-infinity, -4), (-4, 0) and (0, infinity). Choose a test number for each interval:
-6 for the first, -2 for the second and 4 for the third. Determine whether k^2 + 4k > 0 for any of these test numbers:
(-6)^2 + 4(-6) = 36 - 24 = +12. Yes, the discriminant is + on the interval (-infinity, -4)
(-2)^2 + 4(-2) = 4 - 8 = -4. No, the discriminant is not + on (-4, 0).
(4)^2 + 4(4) = 32. Yes, the discrim. is + on (0, infinity).
So: The given quadratic will have two different, real roots on the two intervals (-infinity, -4) and (0, infinity).
