2(n+1)=(3/2)n+1

Solution for 2(n+1)=(3/2)n+1 equation:

2(n+1)=(3/2)n+1

We move all terms to the left:

2(n+1)-((3/2)n+1)=0


Domain of the equation: 2)n+1)!=0

n!=0/1

n!=0

n∈R


We add all the numbers together, and all the variables

2(n+1)-((+3/2)n+1)=0

We multiply parentheses

2n-((+3/2)n+1)+2=0

We multiply all the terms by the denominator


2n*2)n+1)-((+2*2)n+1)+3=0

We add all the numbers together, and all the variables

2n*2)n+1)-(4n+1)+3=0

Wy multiply elements

4n^2=0

a = 4; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·4·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$n=\frac{-b}{2a}=\frac{0}{8}=0$