n(2n+20)+(4n-20)=180

Solution for n(2n+20)+(4n-20)=180 equation:

n(2n+20)+(4n-20)=180

We move all terms to the left:

n(2n+20)+(4n-20)-(180)=0

We multiply parentheses

2n^2+20n+(4n-20)-180=0

We get rid of parentheses

2n^2+20n+4n-20-180=0

We add all the numbers together, and all the variables

2n^2+24n-200=0

a = 2; b = 24; c = -200;
Δ = b2-4ac
Δ = 242-4·2·(-200)
Δ = 2176
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:

$\sqrt{\Delta}=\sqrt{2176}=\sqrt{64*34}=\sqrt{64}*\sqrt{34}=8\sqrt{34}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{34}}{2*2}=\frac{-24-8\sqrt{34}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{34}}{2*2}=\frac{-24+8\sqrt{34}}{4}
$