y=(2m-1)x²-(5m+3)x+3m+5
与x轴相交：y=0
(2m-1)x²-(5m+3)x+3m+5=0
Δ=[-(5m+3)]²-4(2m-1)(3m+5)
=25m²+30m+9-24m²-28m+20
=m²+2m+1+28
=(m+1)²+28
∵无论m取何值,Δ>=28>0
∴无论m取何值,二次函数y=(2m-1)x²-(5m+3)x+3m+5必与x轴有2个交点.