Reversing rsa and obtaining p and q ?

def rfact(n, e, d):
    
    """ because e*d is congruent to 1 mod phi(n), e*d is very close to
      being a multiple of phi, and phi is very close to n for an RSA
      modulus - so the whole number quotient (e*d)/n is 1 less than
      this factor """
    k = 1 + (e * d) / n
    phi = (e * d - 1) / k

    """ if n=p*q, then phi(n) = (p-1)*(q-1) = p*q + 1 - (p+q); but p*q=n,
      so phi(n) = n + 1 - (p+q); set 'sum' to p+q """
    sum = n + 1 - phi

    """ we now know the product of p and q (n) and their sum (sum).  we
      solve for p and q using the quadratic formula.  set 'delta' to
      the radical in the numerator of the formula """
    delta = isqrt(sum**2 - 4*n)

    p = (sum + delta) / 2
    q = (sum - delta) / 2

    if p*q != n:
	print '!Failed:  modulus not factored correctly!'

    return [p, q]

def isqrt(n):
    # construct an initial guess value
    d = len(str(n))	# count of decimal digits
    x = '3'
    d = d / 2
    while d > 0:
        d = d - 1
        x = x + '1'
    x = int(x)

    # perform Newton-Raphson iteration
    prev = 0
    while x != prev:
        prev = x
        q = n / x
        x = (x + q) / 2

    return x