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南清丶丶

#1 Created at ... [Delete] [Delete and Lock User]

-- coding: utf-8 --

print("解方程ax2 + bx + c = 0 (a!=0)")

两边同时除以a得x2+(b/a)x+c/a=0 化简得(x+b/2a)^2=(b^2-4ac)/4a^2

import math

def quadratic(a, b, c): s=(bb-4ac)/(4aa) #等号右边的值 if s<0: return '方程无解' else: x=math.sqrt(s)-(b/2a) m=x #第一个值 n=-x #第二个值 return m,n

a=float(input('请输入参数a')) b=float(input('请输入参数b')) c=float(input('请输入参数c')) if len(quadratic(a, b, c))==2: x1,x2=quadratic(a, b, c) print('x1=',x1,'x2',x2) else: print(quadratic(a, b, c))

m=x #第一个值 n=-x #第二个值 你确定?

星fire燎原

#3 Created at ... [Delete] [Delete and Lock User]

仅仅有提示不安全,如果用户就是要输入a=0


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