320=-8n+4n^2

Solution for 320=-8n+4n^2 equation:

320=-8n+4n^2

We move all terms to the left:

320-(-8n+4n^2)=0

We get rid of parentheses

-4n^2+8n+320=0

a = -4; b = 8; c = +320;
Δ = b2-4ac
Δ = 82-4·(-4)·320
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-72}{2*-4}=\frac{-80}{-8}
=+10
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+72}{2*-4}=\frac{64}{-8}
=-8
$