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斯达叔较难

#1 Created at ... [Delete] [Delete and Lock User] 
#!/usr/bin/env python3
# -*- coding: utf-8 -*-

import math

def quadratic(a, b, c):
    dlt = b ** 2 - 4 * a * c
    sqrtdlt = math.sqrt(dlt)
    if 0 > sqrtdlt:
        return "方程无实数根，但有2个共轭复根"
    elif 0 == sqrtdlt:
        return "方程有两个相等的实数根%s" % (-b / (2 * a))
    elif 0 < sqrtdlt:
        return "方程有两个不相等的实数根%s,%s" % ((-b + sqrtdlt) / (2 * a), (-b - sqrtdlt) / (2 * a))

斯达叔较难

#2 Created at ... [Delete] [Delete and Lock User] 
>>> from fun import quadratic
>>> quadratic(2, 3, 1)
'方程有两个不相等的实数根-0.5,-1.0'
>>> quadratic(1, 3, -4)
'方程有两个不相等的实数根1.0,-4.0'
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