1/n+3=n+2/2

Solution for 1/n+3=n+2/2 equation:

1/n+3=n+2/2

We move all terms to the left:

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


Domain of the equation: n!=0

n∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

1/n-n-1+3=0

We multiply all the terms by the denominator


-n*n-1*n+3*n+1=0

We add all the numbers together, and all the variables

2n-n*n+1=0

Wy multiply elements

-1n^2+2n+1=0

a = -1; b = 2; c = +1;
Δ = b2-4ac
Δ = 22-4·(-1)·1
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{2}}{2*-1}=\frac{-2-2\sqrt{2}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{2}}{2*-1}=\frac{-2+2\sqrt{2}}{-2}
$