You need to use the following factorial formula for arrangements, such that:

`A_n^k = (n!)/((n - k)!)`

You need to re-write each member of equation with respect to the factorial formula, such that:

`A_x^1 = (x!)/((x - 1)!) => A_x^1 = ((x - 1)!*x)/((x - 1)!)`

Reducing duplicate factors yields:

`A_x^1 = x`

`A_(x+1)^2 = ((x + 1)!)/((x + 1 - 2)!)`

`A_(x+1)^2 = ((x + 1)!)/((x - 1)!) => A_(x+1)^2 = ((x - 1)!*x*(x+1))/((x - 1)!) `

Reducing duplicate factors yields:

`A_(x+1)^2 = x(x + 1)`

Replacing x for `A_x^1` and `x(x + 1) ` for `A_(x+1)^2` yields:

`x + x(x + 1) = 24`

Expanding the brackets, yields:

`x + x^2 + x - 24 = 0 => x^2 + 2x = 24`

Adding 1 both sides yields:

`x^2 + 2x + 1 = 24 + 1 => (x + 1)^2 = 25 => x + 1 = +-sqrt 25`

`x + 1 = +-5 => x_1 = 4 ; x_2 = -6`

Since x needs to be a natural number, you cannot validate the value `x = -6.`

**Hence, evaluating the solution to the given equation, using factorial formula of arrangements, yields `x = 4.` **