so the answer for 3 consecutive positive even integers such that the product of the second and third integers is 20 more than ten times the first integer........the answer is n=6 and n=-2

-2 is not allowed, it is NOT a positive integer.

oh? so the answer is just n=6

To find the answer for the question, "What are the three consecutive positive even integers such that the product of the second and third integers is 20 more than ten times the first integer?", we need to follow a step-by-step approach.

Let's assume the first even integer as "n". Since we need three consecutive positive even integers, the next two consecutive even integers can be expressed as "n+2" and "n+4".

Now, according to the given condition, the product of the second and third integers (which is (n+2) * (n+4)) is 20 more than ten times the first integer (which is 10n + 20).

So, we can set up the equation:
(n+2) * (n+4) = 10n + 20

To solve this equation, we can start by expanding the left side:
n^2 + 6n + 8 = 10n + 20

Next, we can simplify by moving all terms to one side:
n^2 - 4n - 12 = 0

Now we have a quadratic equation. To solve for "n", we can factorize it or use the quadratic formula. Factoring in this case gives us:
(n-6)(n+2) = 0

This gives us two possible values. Either "n-6=0" which implies "n=6" or "n+2=0" which implies "n=-2".

Hence, the two possible solutions for "n" are "n=6" and "n=-2". These values correspond to the three consecutive positive even integers, which are:

For n=6: The three consecutive positive even integers are 6, 8, 10.
For n=-2: The three consecutive positive even integers are -2, 0, 2.

Therefore, the answer to the question is n=6 and n=-2.