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缩进很麻烦

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-- coding: utf-8 --

import math

def quadratic(a, b, c): if a!=0: t=(bb)-(4ac) if t<0: return 'eoorr' else: x1=(-b+math.sqrt(t))/(2a) x2=(-b-math.sqrt(t))/(2*a) return x1,x2 else: x1=x2=(-c)/b return x1

廖雪峰

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

缩进确实很麻烦


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