Find three consecutive positive even integers such that the product of the second and third integers is twenty more than ten times the first integer

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You should come up with the following notation for the three consecutive even integers, such that:

`n, n+2, n+4`

The problem provides an information that relates the integers, such that:

`(n+2)(n+4) = 20 + 10n`

`n^2 + 4n + 2n + 8 - 10n - 20 = 0 => n^2 - 4n - 12 = 0`

Using quadratic formula yields:

`n_(1,2) = (4+-sqrt(16 + 48))/2`

`n_(1,2) = (4+-sqrt64)/2 => n_(1,2) = (4+-8)/2`

`n_1 = 6; n_2 = -2`

Since the problem provides the information that the integers are positive, hence, only `n = 6` fulfills the condition.

Hence, evaluating the three consecutive even positive integers, yields 6, 8, 10.

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