$\frac{ 2\left( n+2 \right) \times \left( n+1 \right) ! }{ \left( n+1 \right) ! }=12$
Cancel out the common factor $\left( n+1 \right) !$$2\left( n+2 \right)=12$
Divide both sides of the equation by $2$$n+2=6$
Move the constant to the right-hand side and change its sign$n=6-2$
Subtract the numbers$n=4$
Check if the given value is the solution of the equation$\frac{ 2 \times \left( 4+2 \right) ! }{ \left( 4+1 \right) ! }=12$
Simplify the expression$12=12$
The equality is true, therefore $n=4$ is a solution of the equation$n=4$