Problem:
The sum of the areas of two squares is 544m2, and the difference of their perimeters is 32m. Find the sides of the two squares.

Solution:
Let's denote the side lengths of the two squares as x and y.

Step 1: Set up the equations
We are given two pieces of information:
1. The sum of the areas of the two squares is 544m2:
Area of the first square = x^2
Area of the second square = y^2
x^2 + y^2 = 544 --------(Equation 1)

2. The difference of their perimeters is 32m:
Perimeter of the first square = 4x
Perimeter of the second square = 4y
4x - 4y = 32 --------(Equation 2)

Step 2: Solving the equations
We can solve the above two equations simultaneously to find the values of x and y.

From Equation 2, we can rearrange it as:
4x = 4y + 32
x = y + 8 --------(Equation 3)

Now, substitute the value of x from Equation 3 into Equation 1:
(y + 8)^2 + y^2 = 544

Expanding and simplifying:
y^2 + 16y + 64 + y^2 = 544
2y^2 + 16y + 64 = 544
2y^2 + 16y - 480 = 0

Divide the equation by 2:
y^2 + 8y - 240 = 0

Step 3: Factorizing the quadratic equation
To solve the quadratic equation, we need to factorize it. The factors of -240 that sum up to 8 are 20 and -12.
Therefore, the equation can be factored as:
(y + 20)(y - 12) = 0

Setting each factor equal to zero gives two possible solutions:
y + 20 = 0 or y - 12 = 0

Therefore, y = -20 or y = 12.

We can discard the negative value for y since we are dealing with side lengths, which cannot be negative.

Step 4: Finding the side lengths
Using the value of y = 12, we can substitute it back into Equation 3 to find the value of x:
x = y + 8
x = 12 + 8
x = 20

Therefore, the side lengths of the two squares are 20m and 12m.

Answer:
The sides of the two squares are 20m and 12m.