# ax2 + bx + c = 0. Recall that such an equation has two
#solutions for the value of x, and these are given by the formula
# x = -b +- sqrt(b2 – 4ac) / 2a
import math
a=input("Enter the value of a")
b=input("Enter the value of b")
c=input("Enter the value of c")
d=(b*b-4*a*c)
d=math.sqrt(d) #predefine function in python
root1=(-b+d)/2*a
root2=(-b-d)/2*a
print 'first root is',root1
print '\nsecond root is',root2
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