The length of a rectangular footprint of a residence is 5 m more than its width. If the area of the rectangle is 104m squared, what are the dimensions of the rectangle?
Let's assume the width of the rectangle is x meters.
According to the problem, the length of the rectangle is 5 m more than its width, so the length is x + 5 meters.
The formula to calculate the area of the rectangle is length × width, and it is given that the area is 104 m².
So, (x + 5) × x = 104.
Expanding the equation, we get x² + 5x = 104.
Rearranging the equation, we get x² + 5x - 104 = 0.
This is a quadratic equation that can be factored as (x + 13)(x - 8) = 0.
Setting each factor equal to zero, we get x + 13 = 0 or x - 8 = 0.
From the first equation, x = -13, which does not make sense in this context since width cannot be negative.
From the second equation, x = 8.
So, the width of the rectangle is 8 meters, and the length is 8 + 5 = 13 meters.
Therefore, the dimensions of the rectangle are 8 meters by 13 meters. Answer: \boxed{8, 13}.