The sum of the areas of two squares is 468 m^2: If the difference of their perimeters is 24 m, find the sides of the two squares.

Let's denote the side of the first square as a and the side of the second square as b.

From the problem, we have two equations:

1. The sum of the areas of the two squares is 468 m^2. The area of a square is given by the side length squared, so we have a^2 + b^2 = 468.

2. The difference of their perimeters is 24 m. The perimeter of a square is given by 4 times the side length, so we have 4a - 4b = 24. Simplifying this equation gives a - b = 6.

Now we have a system of two equations, and we can solve it step by step.

First, let's solve the second equation for a: a = b + 6.

Next, substitute a = b + 6 into the first equation: (b + 6)^2 + b^2 = 468.

Expanding and simplifying gives 2b^2 + 12b + 36 = 468.

Subtract 468 from both sides to set the equation to zero: 2b^2 + 12b - 432 = 0.

Divide through by 2 to simplify: b^2 + 6b - 216 = 0.

This is a quadratic equation in the form of (x - p)(x + q) = 0. The solutions to this equation are p and -q.

Factoring the equation gives (b - 12)(b + 18) = 0.

Setting each factor equal to zero gives the possible solutions for b: b = 12 and b = -18.

Since a side length can't be negative, we discard b = -18. So, b = 12.

Substitute b = 12 into the equation a = b + 6 to find a: a = 12 + 6 = 18.

So, the sides of the two squares are 18 m and 12 m.