Let the integers be x and x + 2, then
x(x + 2) = 27 + 6(x + x + 2) = 27 + 6(2x + 2) = 27 + 12x + 12 = 39 + 12x
x^2 + 2x = 39 + 12x
x^2 + 2x - 12x - 39 = 0
x^2 - 10x - 39 = 0
(x - 13)(x + 3) = 0
x - 13 = 0 or x + 3 = 0
x = 13 or x = -3
Therefore the two integers are 13 and 15 or -3 and -1
