n(4+n)=165

Solution for n(4+n)=165 equation:

n(4+n)=165

We move all terms to the left:

n(4+n)-(165)=0

We add all the numbers together, and all the variables

n(n+4)-165=0

We multiply parentheses

n^2+4n-165=0

a = 1; b = 4; c = -165;
Δ = b2-4ac
Δ = 42-4·1·(-165)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-26}{2*1}=\frac{-30}{2}
=-15
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+26}{2*1}=\frac{22}{2}
=11
$