Để pt có 2 nghiệm:
$\Delta'= m^2-m^2+m-3>0$
$\Leftrightarrow m>3$
Theo Viet:
$x_1+x_2=2m$
$x_1x_2= m^2-m+3$
$x_1^2+x_2^2= (x_1+x_2)^2-2x_1x_2= 4m^2-2(m^2-m+3)= 2m^2+2m-6= (\sqrt{2}m)^2 + 2.\sqrt{2}.m.\frac{1}{\sqrt{2}}+ \frac{1}{2}-\frac{13}{2}= (\sqrt{2}m+\frac{1}{\sqrt{2}} )^2-\frac{13}{2}\ge \frac{-13}{2}$
$\Rightarrow min=\frac{-13}{2}$