Firstly, because this equation is solvable in real numbers, therefore $$(2m)^2-4(m^2+2m+3)\ge 0\implies x\le -\frac{3}{2} $$ By the Vieta's theorem, we have $$x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=(2m)^2-2(m^2+2m+3)=2m^2-4m-8$$ Hence, this problem is equivalent to finding the minimal value of the above quadratic function when $m\le -\frac{3}{2}$. Write the target as $$2m^2-4m-6 = 2(m-1)^2 - 8$$ Hence we find when $m=-\frac{3}{2}$, the expression reaches minimal at $\boxed{\frac{9}{2}}$.