Question 397738
Let x, x+2, x+4 be the integers, where x is even. Then,


{{{(x+2)(x+4) = 20 + 10x}}}


{{{x^2 + 6x + 8 = 20 + 10x}}}


{{{x^2 - 4x - 12 = 0}}}


{{{(x-6)(x+2) = 0}}}


x = 6, -2


Checking, both of these satisfy, but the question looks for positive even integers. So {6, 8, 10} are the three integers.