求一元二次方程：ax^2+bx+c=0的解，為啥不對
import math
def quadratic_equation(a, b, c):
??? tmp = b * b - 4 * a * c
??? if tmp < 0:
??????? return
??? elif tmp == 0:
??????? return -b/(2 * a)
??? else:
??? x1= (-b + math.sqrt(tmp))/(2*a)
??? x2= (-b - math.sqrt(tmp))/(2*a)
??? return x1,x2
print quadratic_equation(2, 3, 0)
print quadratic_equation(1, -6, 5)
2016-11-09
import math
def quadratic_equation(a, b, c):
? ? tmp = b * b - 4 * a * c?
? ? if tmp < 0:
? ? ? ? return?
? ? elif tmp == 0:
? ? ? ? return -b/(2 * a)
? ? else:
? ? ? ? x1= (-b + math.sqrt(tmp))/(2*a)
? ? ? ? x2= (-b - math.sqrt(tmp))/(2*a)
? ? ? ? return x1,x2
print quadratic_equation(2, 3, 0)
print quadratic_equation(1, -6, 5)
代碼的邏輯沒有問題，出的bug是縮進問題，把else后面的代碼進行縮進就可以了