You need to use factorial formula of arrangements of `n` elements taken `k` at a time such that:

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

Substituting `2x + 1` for `n` and `2` for `k` yields:

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

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

You may write `(2x+1)!` as `(2x-1)!(2x)(2x+1)` such that:

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

Reducing like terms yields:

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

Substituting `2x(2x+1)` for `A_(2x+1)^2` in equation yields:

`2x(2x+1) = 110`

Reducing by 2 both sides yields:

`x(2x+1) = 55`

You need to open brackets such that:

`2x^2 + x = 55 =gt 2x^2 + x- 55 = 0`

You need to use quadratic formula such that:

`x_(1,2) = (-1+-sqrt(1+440))/4 =gt x_(1,2) = (-1+-21)/4`

`x_1 = 5 ; x_2 = -22/4 =gt x_2 = -11/2`

**You need to keep only the natural number `5` , hence the solution to the equation is `x=5` .**