2+3n^2/4n=4

Solution for 2+3n^2/4n=4 equation:

2+3n^2/4n=4

We move all terms to the left:

2+3n^2/4n-(4)=0


Domain of the equation: 4n!=0

n!=0/4

n!=0

n∈R


We add all the numbers together, and all the variables

3n^2/4n-2=0

We multiply all the terms by the denominator


3n^2-2*4n=0

Wy multiply elements

3n^2-8n=0

a = 3; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·3·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*3}=\frac{0}{6}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*3}=\frac{16}{6}
=2+2/3
$