The sum of the squares of the three consecutive even natural numbers is 1460. Find the numbers.

Let's denote the three consecutive even natural numbers as n, n+2, and n+4.

The problem states that the sum of the squares of these numbers is 1460. So, we can write the equation as:

n^2 + (n+2)^2 + (n+4)^2 = 1460

Expanding and simplifying this equation gives:

3n^2 + 12n + 20 = 1460

Subtract 1460 from both sides to set the equation to zero:

3n^2 + 12n + 20 - 1460 = 0

Simplify to get:

3n^2 + 12n - 1440 = 0

Divide the entire equation by 3 to simplify further:

n^2 + 4n - 480 = 0

This is a quadratic equation in the form of ax^2 + bx + c = 0. We can solve for n using the quadratic formula, n = [-b ± sqrt(b^2 - 4ac)] / (2a):

n = [-4 ± sqrt((4)^2 - 41(-480))] / (2*1) n = [-4 ± sqrt(16 + 1920)] / 2 n = [-4 ± sqrt(1936)] / 2 n = [-4 ± 44] / 2 n = 20 or n = -24

Since n represents a natural number, we discard -24 as a solution. Therefore, the three consecutive even natural numbers are 20, 22, and 24.