y=2x^2+kx+3k-4的图像和x轴的两个交点的距离是6那么设2x^2+kx+3k-4=0的两根是x1,x2(x1＜x2)故x2-x1=6由韦达定理有x1+x2=-k/2,x1*x2=(3k-4)/2所以(x2-x1)^2=(x1+x2)^2-4x1*x2=(-k/2)^2-4*(3k-4)/2=36即k^2-24k-112=0...