Python数学实现二元一次方程

 import cmath
 import  math
 import  sys

 def get_float(msg,allow_zero):
     x = None
     while x is None:
         try:
             x = float(input(msg))
             if not allow_zero and abs(x) <sys.float_info.epsilon:
                 print("zero is not allowed")
                 x = None
         except ValueError as err:
             print(err)
     return x

 print("ax\N{SUPERSCRIPT TWO} +bx +c = 0")
 a = get_float("enter a:", False)
 b = get_float("enter b:", True)
 c = get_float("enter c:", True)
 x1 = None
 x2 = None
 discriminant = (b**2)-(4*a*c)
 if discriminant == 0:
     x1 = -(b/(2*a))
 else:
     if discriminant >0:
         root  = math.sqrt(discriminant)
     else:
         root = cmath.sqrt(discriminant)

     x1 = (-b+root) / (2*a)
     x2 = (-b -root) /(2*a)

 equation = ("{0}x\N{SUPERSCRIPT TWO}+{1}+{2} = 0""\N{RIGHTWARDS ARROW}X = {3}").format(a,b,c,x1)
 if x2 is not None:
     equation += "  or x = {0}".format(x2)
 print(equation)