achloes
IanStewart ScottTargetTestPrep Thanks for your solutions - very helpful. I do have a follow up question though.
If k is 90, isn't the other root -15? In which case, how do we end up with a 3 for the middle term?
My work below:
2x^2 + 3x - 90 = 0
(2x - 6) (2x + 15) = 0
Adding -6 and 15 gives me 9, not 3. Really don't know where I'm going wrong

I think you are referring to the rule that the constant term in a quadratic equation in the standard form is the product of the roots, but remember that it only works for quadratics of the form x^2 + bx + c = 0, where the coefficient of the x^2 term is 1.
For a general quadratic of the form ax^2 + bx + c = 0, the product of the roots is c/a, so the product of the roots of the equation 2x^2 + 3x - 90 = 0 is c/a = (-90)/2 = -45, which means that the other root of the equation is -45/6 = -15/2. So you can rewrite the quadratic in the factored form as a(x - p)(x - q) = 2(x - 6)(x + 15/2) = (x - 6)(2x + 15). Also, the sum of the roots of an equation of the form ax^2 + bx + c = 0 is -b/a, which means the sum of the roots of 2x^2 + 3x - 90 = 0 is -3/2. We can verify that the sum of the roots is indeed 6 + (-15/2) = -3/2.