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