[编程入门]自定义函数求一元二次方程-题解（Python代码）

解题思路:
见代码

参考代码:

def x0(a,b,c):
    s = b**2-4*a*c
    if s > 0:
        x1 = -b+(s)**(1/2)/(2*a)
        x2 = -b-(s)**(1/2)/(2*a)
        print('x1={:.3f} x2={:.3f}'.format(x1,x2))
    elif s == 0:
        x1=x2=-b/2*a
        print('x1={:.3f} x2={:.3f}'.format(x1,x2))
    elif s < 0:
        l = ((s**(1/2))/(2*a)).imag   #取出a+bj中b的值
        print('x1={0:.3f}+{1:.3f}i x2={0:.3f}-{1:.3f}i'.format(-b/(2*a),l))

a,b,c = input().split()
a,b,c = int(a),int(b),int(c)
x0(a,b,c)