Question

(4-y^2)/(y-2) = y/2

(4-y^2)/(y-2) = y/2

4 - y^2 = (y*(y-2))/2

4 - y^2 = (y^2-2y)/2

8 - 2y^2 = y^2 - 2y

3y^2 - 2y - 8 = 0
(keep the above equation in mind)

quadratic formula:
(-b +/- sqrt(b^2 - 4ac))/2a

now solve by plugging in the appropriate numbers and choose (normally) the positive or higher value.

(4-y^2)/(y-2) = y/2 Factor
(y-2)(y+2)/(Y-2)= y/2
y+2=y/2
y/2=-2
y=-4

Julie made and error in her post. She multiplied both sides by y-2. This automatically excludes y from being 2. In here conclusion to then proceed, you will get y=2 as a solution in the quadratic. It is not valid.
One cannot multiply both sides of an equation by zero and get a valid answer.