/(n)=3n-4

Solution for /(n)=3n-4 equation:

/(n)=3n-4

We move all terms to the left:

/(n)-(3n-4)=0


Domain of the equation: n!=0

n∈R


We get rid of parentheses

/n-3n+4=0

We multiply all the terms by the denominator


-3n*n+4*n+=0

We add all the numbers together, and all the variables

4n-3n*n=0

Wy multiply elements

-3n^2+4n=0

a = -3; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-3)·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*-3}=\frac{-8}{-6}
=1+1/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-3}=\frac{0}{-6}
=0
$