SOLUTION: I need to find the product of two consecutive odd integers is 2 less than six times their sum. Can you find the two integers. do I set it up like this and is this the correct p

Question 268921: I need to find the product of two consecutive odd integers is 2 less than six times their sum. Can you find the two integers.
do I set it up like this and is this the correct process to solve:
n(n+2) = 6(n+n+2)-2
n^2 + 2n = 6(2n + 2 - 2)
n^2 + 2n = 12n + 12 - 12
n^2 + 2n -12n = -12n + 12n + 12 - 12
n^2 - 10n = 12 - 12
n^2 - 10n + 12 = 12 - 12 + 12
n^2 - 10n + 12 = 12
and this is were I get stuck. Do I remove -12 from both sides? If I do I know I will get n^2 - 10n - 0 = 0 is that correct and if not what am I doing wrong?

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
I need to find the product of two consecutive odd integers is 2 less than six times their sum. Can you find the two integers.
do I set it up like this and is this the correct process to solve:
-------------------
n(n+2) = 6(n+n+2)-2
n^2 + 2n = 6(2n + 2 - 2) *** = 6(2n+2) - 2
n^2 + 2n = 12n + 12 - 2
n^2 + 2n = 12n + 10
n^2 - 10n - 10 = 0
-------------------
There are no integer solutions. The product of 2 odd numbers is odd. 6 times their sum is even, and subtracting 2 is still even.
---> no solutions
You made one error, but it didn't matter, there are no such numbers.
-----------------
n^2 + 2n = 12n + 12 - 12
n^2 + 2n -12n = -12n + 12n + 12 - 12
n^2 - 10n = 12 - 12
n^2 - 10n + 12 = 12 - 12 + 12
n^2 - 10n + 12 = 12

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