键盘输入一元二次方程（ax2+bx+c=0）的二次项到常项系数a，b，c，判断方程是否有实根，如有实根则保留两位小数输出，如没有实根则输出提示信息

请输入二次项系数a：1
请输入一次项系数b：6
请输入常数项c：3
该一元二次方程有两个实数根，分别是
x1=-0.55*2=-5.45

from numpy.lib.scimath import sqrt

a = int(input())
b = int(input())
c = int(input())
if a == 0:
        if b == 0 and c ==0:
            print("There are infinitely many solutions!")
        else:
            if b==0:
                print ("There is no solution")
            else:
                print("There is a unique solution x=%g" %(-1*c/b))
else:
    disc = b*b -4*a*c
    if disc > 0:
        x1 = (-b + sqrt(disc)) / 2 / a
        x2 = (-b - sqrt(disc)) / 2 / a
        if disc > 0:
            print("There are two distinct real roots x1=%g and x2=%g" % (x1, x2))
        else:
            print("There are two conjugate roots x1=%g and x2=%g" % (x1, x2))
    else:
        if disc == 0:
            print("There are two equal real roots x1=x2=%g" % (-1 * b / 2 / a))
        else:
            x1 = (-b + sqrt(disc)) / 2 / a
            x2 = (-b - sqrt(disc)) / 2 / a
            print("There are two conjugate roots x1=%s and x2=%s" % (x1, x2))

如有帮助，采纳支持一下，谢谢。

import math

def is_real_root(a,b,c):
  return (b * b - 4 * a * c) < 0

def get_real_root(a,b,c):
  if(a == 0):
    print('此方程不是一元二次方程！')
  elif(is_real_root(a,b,c)):
    print('此方程一元二次方程无实数根！')
  else:
    x1 = (-b + math.sqrt(b * b - 4 * a * c))/(2 * a)
    x2 = (-b - math.sqrt(b * b - 4 * a * c))/(2 * a)
    print(f'该一元二次方程有两个实数根，分别是:x1={round(x1,2)},x2={round(x2,2)}')


if __name__ == '__main__':
  a = int(input('请输入二次项系数a：'))
  b = int(input('请输入一次项系数b：'))
  c = int(input('请输入常数项c：'))
  get_real_root(a,b,c)