当k=0时,原方程即 x-2=0解得x=2
当k≠0时,
Δ=（3k-1)^2-8k(k-1)
=k^2+2k+1
=(k+1)^2≥0
x1,2=[(3k-1)±(k+1)]/(2k)
x1=(k-1)/k, x2=2