-n+(3/2)n=0

Solution for -n+(3/2)n=0 equation:

-n+(3/2)n=0


Domain of the equation: 2)n!=0

n!=0/1

n!=0

n∈R


We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses

3n^2-1n=0

a = 3; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·3·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*3}=\frac{0}{6}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*3}=\frac{2}{6}
=1/3
$