n(n+2)=86+(n+4)

Solution for n(n+2)=86+(n+4) equation:

n(n+2)=86+(n+4)

We move all terms to the left:

n(n+2)-(86+(n+4))=0

We multiply parentheses

n^2+2n-(86+(n+4))=0


We calculate terms in parentheses: -(86+(n+4)), so:

86+(n+4)

determiningTheFunctionDomain
(n+4)+86

We get rid of parentheses

n+4+86

We add all the numbers together, and all the variables

n+90

Back to the equation:

-(n+90)


We get rid of parentheses

n^2+2n-n-90=0

We add all the numbers together, and all the variables

n^2+n-90=0

a = 1; b = 1; c = -90;
Δ = b2-4ac
Δ = 12-4·1·(-90)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-19}{2*1}=\frac{-20}{2}
=-10
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*1}=\frac{18}{2}
=9
$