b^2-4ac=(m+2)^2-4*3*(m^2+1) =m^2+4m+4-12m^2-12 =-11m^2+4m-8 =-11[m^2-(4/11)m+4/121]+(4/121-8) =-11(m-2/11)^2-964/121 <0 即b^2-4ac<0所以方程没有实根.