Help with algebra question

Find 3 positive even integers such that the produnt of the first and second is 8 more than 38 times the third. I've tried doing this but i dont get it. Please explain

Three positive even integers can be represented as (N - 2), N and (N + 2).

The product of the first two is (N - 2) * N or N^2 - 2N

Since that's 8 more than 38 times the third, add 8 to the product of 38 and (N + 2).

Equate the two:

N^2 - 2N = [38 * (N + 2)] + 8

Simplify the right side:

N^2 - 2N = 38N + 76 + 8 = 38N + 84

Subtract 38N and 84 from both sides:

N^2 - 40N - 84 = 0

Factor to:

(N - 42)(N + 2) = 0

Ignore the -2 negative root.

N is 42. That makes the three numbers 40, 42, and 44.

a lot of the answerers on here are ASSUMING they need to be consecutive even integers, but this is not writtem: are there other solutions if we lift this false constraint? Namely: we need to generate integer answers to the question.
I thought that by starting with A = 2a, B=2b, and C = 2c we could insure that the answers are even but not necessarily consecutive.

Following this logic, I get to ab = 19c + 2, which seems to indicate many solutions, but with the constraints of integral answers imposed. How do we do that/find ALL of the possible solutions from here?
for example: A=8, B=20, C=4 work here, selecting C first so that we are not humbled by divisibility of 38! also working are (4,40,4) and (2,80,4) (20,8,4) among MANY others, as we can set C to be any even positive integer and then solve for the even factors of 38C+8

Let the even integers be x-2;x;x+2
now product of first and second is x^2-2x
and 38 times the third is 38(x+2).
now if you subtract the second expression from the first you get 8
so
x^2-2x-38(x+2)=8
x^2-40x-76=8
x^2-40x-84=0
x^2-42x+2x-84=0
x(x-42)+2(x-42)=0
(x-42)(x+2)=0