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

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

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

We move all terms to the left:

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


Domain of the equation: 3)(1+n)!=0

n∈R



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

n!=0/1

n!=0

n∈R


We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We calculate fractions


(2n^2+4n)/6n^2+()/6n^2=0

We multiply all the terms by the denominator


(2n^2+4n)+()=0

We add all the numbers together, and all the variables

(2n^2+4n)=0

We get rid of parentheses

2n^2+4n=0

a = 2; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·2·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*2}=\frac{-8}{4}
=-2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*2}=\frac{0}{4}
=0
$