∵△=(2m+3)-4(m-1) =(2m+3+2m-2)(2m+3-2m+2) =5(4m+1) 又m<-1/4,则4m+1<0 ∴△=5(4m+1)<0 故方程x^2+(2m+3)x+(m-1)^2=0无实根.