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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2n^2+4n=0

a = -2; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-2)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-2}=\frac{-8}{-4}
=+2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-2}=\frac{0}{-4}
=0
$