3+20/n2=13

Solution for 3+20/n2=13 equation:

3+20/n2=13

We move all terms to the left:

3+20/n2-(13)=0


Domain of the equation: n2!=0

n^2!=0/

n^2!=√0

n!=0

n∈R


We add all the numbers together, and all the variables

20/n2-10=0

We multiply all the terms by the denominator


-10*n2+20=0

We add all the numbers together, and all the variables

-10n^2+20=0

a = -10; b = 0; c = +20;
Δ = b2-4ac
Δ = 02-4·(-10)·20
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{2}}{2*-10}=\frac{0-20\sqrt{2}}{-20}
=-\frac{20\sqrt{2}}{-20}
=-\frac{\sqrt{2}}{-1}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{2}}{2*-10}=\frac{0+20\sqrt{2}}{-20}
=\frac{20\sqrt{2}}{-20}
=\frac{\sqrt{2}}{-1}
$