if (b**2-4*a*c)<0:
print('此函数无解')
else:
x1 = (-b+math.sqrt(b**2-4*a*c))/(2*a)
x2 = (-b-math.sqrt(b**2-4*a*c))/(2*a)
return x1,x2
也不能等于0
也可以吧，小于零无解，等于零两个解的值相同，大于零，两个解的值不同，我觉得这样也行吧
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跳跃天空020
if (b**2-4*a*c)<0:
print('此函数无解')
else:
x1 = (-b+math.sqrt(b**2-4*a*c))/(2*a)
x2 = (-b-math.sqrt(b**2-4*a*c))/(2*a)
return x1,x2