Let the two integers be $x$ and $y$, where $x>y$.
We are given that $x-y=6$ and $xy=112$.
We can rewrite the first equation as $x=y+6$ and substitute into the second equation to get $(y+6)y=112$.
Simplifying, we have $y^2+6y-112=0$.
Factoring, we get $(y+14)(y-8)=0$.
So, $y=-14$ or $y=8$.
Since we are looking for positive integers, we discard $y=-14$.
Therefore, $y=8$ and $x=y+6=14$.
The sum of the two integers is $8+14=\boxed{22}$.
The answer is: 22
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Given that the difference between two positive integers is 6 and their product is 112, what is the sum of the integers?

### Answer:
