Find three positive consecutive integers such that the product of the first and second is 2 more than 9 times the third

Three positive consecutive integers are n, n+1, and n+2. And these integers satisfy:

n (n+1) = 9 (n+2) + 2 expanding each side gives you:

n^2+n=9n+18+2

n^2+n=9n+20 subtract 9n from both sides

n^2-8n=20 subtract 20 from both sides

n^2-8n-20=0 now factor

n^2+2n-10n-20=0

n (n+2) - 10 (n+2) = 0

(n-10) (n+2) = 0, since we only want positive integers

n=10

So the three integers are 10, 11, 12

10, 11, and 12 would be ur answer.