jmc

The points of intersection of the line $y=5$ and $y=x^2+3x+2$ are found when $x^2+3x+2=5$. Thus we have the quadratic $x^2+3x-3=0$. By the quadratic formula, $$x=\frac{-3\pm \sqrt{3^2-4\cdot 1\cdot -3}}{2\cdot 1}=\frac{-3\pm \sqrt{21}}{2}$$We want to find the difference of these roots to find the difference of the x-coordinates of the points of intersection, which will give a side length of the square. The difference is $2\cdot \frac{\sqrt{21}}{2}=\sqrt{21}$.

Therefore, the area of the square is the square of the side length, which is $(\sqrt{21}{)}^2=\boxed{21}$.