1028: [编程入门]自定义函数求一元二次方程

解题思路:

注意事项:

参考代码:

import math

a,b,c = map(int,input().split())

delt = b**2-4*a*c

if delt > 0:

    print("x1={:.3f} x2={:.3f}".format((-b+math.sqrt(delt))/(2*a),(-b-math.sqrt(delt))/(2*a)))

elif delt == 0:

    print("x1={:.3f} x2={:.3f}".format(-b/(2*a),-b/(2*a)))

else:

    print("x1={:.3f}+{:.3f}i x2={:.3f}-{:.3f}i".format(-b/(2*a),math.sqrt(-delt)/(2*a),-b/(2*a),math.sqrt(-delt)/(2*a)))