The longest distance

$\text {The distance between the curve } y = \sqrt {3+2x-2x^2}$ ) and the point O(0,0):

$D = \sqrt {(x - 0)^2 + (y - 0)^2} = \sqrt {x^2 + y^2} = \sqrt {x^2 + (3+2x-2x^2)} =$

$= \sqrt {3 + 2x - x^2)} = \sqrt {4 - (x - 1)^2}$

From the last formula, it is easy to see, that D is maximal when x = 1 ( and $y = \sqrt {3}$ ), and the maximum distance:

$D = \sqrt {4 - (1 - 1)^2} = \sqrt {4 - 0^2} = \boxed {2}$