n2+6=(3+n)2

Solution for n2+6=(3+n)2 equation:

n2+6=(3+n)2

We move all terms to the left:

n2+6-((3+n)2)=0

We add all the numbers together, and all the variables

n2-((n+3)2)+6=0

We add all the numbers together, and all the variables

n^2-((n+3)2)+6=0


We calculate terms in parentheses: -((n+3)2), so:

(n+3)2

We multiply parentheses

2n+6

Back to the equation:

-(2n+6)


We get rid of parentheses

n^2-2n-6+6=0

We add all the numbers together, and all the variables

n^2-2n=0

a = 1; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·1·0
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{-(-2)-2}{2*1}=\frac{0}{2}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*1}=\frac{4}{2}
=2
$