N*n=320/n

Solution for N*n=320/n equation:

*N=320/N

We move all terms to the left:

*N-(320/N)=0


Domain of the equation: N)!=0

N!=0/1

N!=0

N∈R


We add all the numbers together, and all the variables

*N-(+320/N)=0

We add all the numbers together, and all the variables

N-(+320/N)=0

We get rid of parentheses

N-320/N=0

We multiply all the terms by the denominator


N*N-320=0

Wy multiply elements

N^2-320=0

a = 1; b = 0; c = -320;
Δ = b2-4ac
Δ = 02-4·1·(-320)
Δ = 1280
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{5}}{2*1}=\frac{0-16\sqrt{5}}{2}
=-\frac{16\sqrt{5}}{2}
=-8\sqrt{5}
$
$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{5}}{2*1}=\frac{0+16\sqrt{5}}{2}
=\frac{16\sqrt{5}}{2}
=8\sqrt{5}
$