由题意得：
△=3k+1-4(2k-a)
=3k+1-8k+4a>=0
则k<=(4a+1)/5
又3k+1>=0,
则k>=-1/3
若(4a+1)/5<-1/3,即a<-2/3
k不存在.
若(4a+1)/5=-1/3,即a=-2/3
k=-1/3
若(4a+1)/5=-1/3,即a>-2/3
-1/3<=k<=(4a+1)/5
综上：当a<-2/3时,k不存在.
当a=-2/3时,k=-1/3
当a>-2/3时,k属于[-1/3,(4a+1)/5]