Use Heaviside's method to find \(A\).
\[ \dfrac{3x^2+7x+10}{(x+2)(x^2+x+2)}\cdot(x+2) = \dfrac{3x^2+7x+10}{x^2+x+2} \] \[ A=\dfrac{3x^2+7x+10}{x^2+x+2}\Huge|_{\normalsize x=-2}\normalsize=2 \]Therefore
\[ \dfrac{3x^2+7x+10}{(x+2)(x^2+x+2)}=\dfrac{2}{x+2}+\dfrac{Bx+C}{x^2+x+2} \]Clearing fractions yields
\[ 3x^2+7x+10=2(x^2+x+2)+(x+2)(Bx+C)=(2+B)x^2+(2+2B+C)x+(2C+4) \]Therefore, \(2+B=3,\quad 2+2B+C=7,\quad 2C+4=10\). From which we deduce that \(B=1\) and \(C=3\).
So the decomposition is
\[ \dfrac{3x^2+7x+10}{(x+2)(x^2+x+2)}=\dfrac{2}{x+2}+\dfrac{x+3}{x^2+x+2} \]