If y=x^2+kx-k, for what values of k will the quadratic have two real solutions

We use the Factor and Remainder Theorem to answer this problem. Suppose you have a polynomial in terms of x. When you substitute a value x and the answer would be zero, then that means that x is a factor of the polynomial producing 2 real solutions. If the answer is nonzero, then it is not a factor. Therefore, we also have to know at what value of x should we use. Let's assume x-2. Applying the factor and remainder theorem:

y = x ² + kx-k = 0

(2) ² + k (2) - k = 0

k = - 4