You can put this solution on YOUR website! To find the area enclosed, we have to determine the points of intersection...
Turns out they intersect at x = -2 and x = 3...thus we integrate from -2 to 3...
Area = Integral[(x=-2) to (=3)] [(x^2 + 2x + 4)-(2x^2 + x - 2)]
Area = Integral[(x=-2) to (=3)] [(x^3/3 + x^2 + 4x)-((2/3)x^3 + (1/2)x^2 - 2x))]
Area = -x^3/3 + (3/2)x^2 + 6x
Now evaluate from -2 to 3...we get
Area = 125/6