n2=(1+n)2+5

Solution for n2=(1+n)2+5 equation:

n2=(1+n)2+5

We move all terms to the left:

n2-((1+n)2+5)=0

We add all the numbers together, and all the variables

n2-((n+1)2+5)=0

We add all the numbers together, and all the variables

n^2-((n+1)2+5)=0


We calculate terms in parentheses: -((n+1)2+5), so:

(n+1)2+5

We multiply parentheses

2n+2+5

We add all the numbers together, and all the variables

2n+7

Back to the equation:

-(2n+7)


We get rid of parentheses

n^2-2n-7=0

a = 1; b = -2; c = -7;
Δ = b2-4ac
Δ = -22-4·1·(-7)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4\sqrt{2}}{2*1}=\frac{2-4\sqrt{2}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4\sqrt{2}}{2*1}=\frac{2+4\sqrt{2}}{2}
$