Question

How do you solve `-x^2+2x>=0`?

Answer 1

`0<=x<=2`.
See graphical depiction, for both 1-D {(x)} and 2-D {(x, y)}

Explanation:

`-x^2+2x=-(x-1)^2+1>=0`. So,

`-(x-1)^2>=-1 to (x-1)^2<=1`, And so,

`|x-1|<=1.`

Thus, `-1<=x-1<=1`, giving

`0<=x<=2`

graph{-x^2+2x>=0 [-2, 2, -1, 1]}