3y^2+2y-1=43y^2+2y-5=0(3y+5)(y-1)=0 （十字相乘法）故y1=-5/3、y2=1
解(3y-1)(y+1)=43y^2+3y-y-1=43y^2+2y-5=0(3y+5)(y-1)=0∴y=1或y=-5/3
3y^2+2y-5=0(y-1)(3y+5)=0y1=1 y2=-5/3
