给定一个二次方程，任务是找到它的可能解。
例子：
Input :
enter the coef of x2 : 1
enter the coef of x : 2
enter the costant : 1
Output :
the value for x is -1.0
Input :
enter the coef of x2 : 2
enter the coef of x : 3
enter the costant : 2
Output :
x1 = -3+5.656854249492381i/4 and x2 = -3-5.656854249492381i/4
算法 ：
Start.
Prompt the values for a, b, c.
Compute i = b**2-4*a*c
If i get negative value g=square root(-i)
Else h = sqrt(i)
Compute e = -b+h/(2*a)
Compute f = -b-h/(2*a)
If condition e==f then
Print e
Else
Print e and f
If i is negative then
Print -b+g/(2*a) and -b-g/(2*a)
stop
以下是上述任务的Python实现。
# Python program for solving a quadratic equation.
from math import sqrt
try:
# if user gives non int values it will go to except block
a = 1
b = 2
c = 1
i = b**2-4 * a * c
# magic condition for complex values
g = sqrt(-i)
try:
d = sqrt(i)
# two resultants
e = (-b + d) / 2 * a
f = (-b-d) / 2 * a
if e == f:
print("the values for x is " + str(e))
else:
print("the value for x1 is " + str(e) +
" and x2 is " + str(f))
except ValueError:
print("the result for your equation is in complex")
# to print complex resultants.
print("x1 = " + str(-b) + "+" + str(g) + "i/" + str(2 * a) +
" and x2 = " + str(-b) + "-" + str(g) + "i/" +
str(2 * a))
except ValueError:
print("enter a number not a string or char")
输出 ：
the values for x is -1.0解释 ：
首先，该程序将从用户那里获得三个输入。值是
，系数
并保持不变。然后执行公式
对于复杂的值
变得消极。生根为负值将引发值错误。在这种情况下，将结果
然后将其植根。别忘了包括
最后。