n(n+2)+(n+4)=84

Solution for n(n+2)+(n+4)=84 equation:

n(n+2)+(n+4)=84

We move all terms to the left:

n(n+2)+(n+4)-(84)=0

We multiply parentheses

n^2+2n+(n+4)-84=0

We get rid of parentheses

n^2+2n+n+4-84=0

We add all the numbers together, and all the variables

n^2+3n-80=0

a = 1; b = 3; c = -80;
Δ = b2-4ac
Δ = 32-4·1·(-80)
Δ = 329
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{329}}{2*1}=\frac{-3-\sqrt{329}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{329}}{2*1}=\frac{-3+\sqrt{329}}{2}
$