smc

There are three cases: (i) $(x+2),(x-2)\ge 0$ (ii) $x+2\ge 0,x-2<0$ (iii) $(x+2),(x-2)<0$ In case (i), we solve the equation $x+2=2(x-2)$ to get $x=6$. In case (ii), we solve the equation $x+2=2(2-x)$ to get $x=\frac{2}{3}$. Case (iii) simply produces an equation which is that of case (i) multiplied on both sides by $-1$, so it yields the same solutions and thus may be discarded. Thus, if we add the real values of $x$ which satisfy the original equation, we get $6+\frac{2}{3}=6\frac{2}{3}$, or answer choice $\boxed{6\frac{2}{3}}$.