what is solution of equation arrangements(x+1,2)=combinations(x-1,2)+12?

A(x+1,2) = C(x-1,2)+12

We'll
recall the formula of arrangements:

A(x+1,2) =
(x+1)!/(x+1-2)! =  (x+1)!/(x-1)!

But (x+1)! =
(x-1)!*x*(x+1) => (x+1)!/(x-1)! =
(x-1)!*x*(x+1)/(x-1)!

We'll reduce like
terms:

A(x+1,2) =
x*(x+1)

We'll recall the formula of
combinations:

C(x-1,2) = (x-1)!/2!*(x-1-2)!
=(x-1)!/2!*(x-3)!

But (x-1)! = (x-3)!*(x-2)*(x-1)
=>C(x-1,2) = (x-3)!*(x-2)*(x-1)/2!*(x-3)!

But 2! =
1*2 = 2

We'll reduce like
terms:

C(x-1,2) =
(x-2)*(x-1)/2

We'll re-write the
equation:

A(x+1,2) = C(x-1,2)+12 <=>
x*(x+1)=(x-2)*(x-1)/2 + 12

We'll remove the brackets and
we'll multiply by 2 both sides:

2x^2 + 2x = x^2 - 3x + 2 +
24

We'll move all terms to the left
side:

x^2 + 5x - 26 = 0

We'll
apply quadratic formula:

x1 =
(-5+sqrt(25+104))/2

Since sqrt129 is an irrational number,
then the solutions of the equation are irrational, which is impossible because x must be
a natural number.

Therefore, the given
equation has no natural solutions.