2n+4n(n-2)=8+8n

Solution for 2n+4n(n-2)=8+8n equation:

2n+4n(n-2)=8+8n

We move all terms to the left:

2n+4n(n-2)-(8+8n)=0

We add all the numbers together, and all the variables

2n+4n(n-2)-(8n+8)=0

We multiply parentheses

4n^2+2n-8n-(8n+8)=0

We get rid of parentheses

4n^2+2n-8n-8n-8=0

We add all the numbers together, and all the variables

4n^2-14n-8=0

a = 4; b = -14; c = -8;
Δ = b2-4ac
Δ = -142-4·4·(-8)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-18}{2*4}=\frac{-4}{8}
=-1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+18}{2*4}=\frac{32}{8}
=4
$