Find two consecutive even integers such that 5 times their sum is 10 less than their product. (Enter your answers as a comma-separated list.)

Let the lower number be n. Then

5*(n + n+ 2) = n*(n + 2) - 10 10n + 10 = n^2 +2n -10

Rearrange

n^2 - 8n - 20 = 0

This factors as (n + 2)*(n - 10) = 0

I'll let you take it from here.