1. The problem statement, all variables and given/known data Solve 4y-y^{2}=x for y. 3. The attempt at a solution First I tried using the quadratic equation with a=-1 b=4 c=-x y=(-b±(4^{2}-4(-1)(-x))^{(1/2)})/2(-1) That got me this far: y=2±(8-2x)^{(1/2)} Then I checked wolfram and now I am confused Spoiler y=±2-sqrt(4-x . Did I not approach this corectly?

Your approach is fine. You just made an algebra mistake while simplifying. You had ##y = \frac{-4 \pm \sqrt{16 - 4x}}{-2}##. Note that you can't simply divide the stuff inside the radical by 2 when you bring the 2 from the denominator inside.

Thank you for the quick reply and sorry for my slow thanks. How very typical of me that it's some algebraic error Thanks again!