(n+0.1)2+(2n)2=(n+0.6)2+(n)2

Solution for (n+0.1)2+(2n)2=(n+0.6)2+(n)2 equation:

(n+0.1)2+(2n)2=(n+0.6)2+(n)2

We move all terms to the left:

(n+0.1)2+(2n)2-((n+0.6)2+(n)2)=0

We add all the numbers together, and all the variables

2n^2+(n+0.1)2-((n+0.6)2+n2)=0

We multiply parentheses

2n^2+2n-((n+0.6)2+n2)+0.2=0


We calculate terms in parentheses: -((n+0.6)2+n2), so:

(n+0.6)2+n2

We add all the numbers together, and all the variables

n^2+(n+0.6)2

We multiply parentheses

n^2+2n+1.2

Back to the equation:

-(n^2+2n+1.2)


We get rid of parentheses

2n^2-n^2+2n-2n-1.2+0.2=0

We add all the numbers together, and all the variables

n^2-1=0

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

$\sqrt{\Delta}=\sqrt{4}=2$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2}{2*1}=\frac{-2}{2}
=-1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2}{2*1}=\frac{2}{2}
=1
$