0=-2n^2+40n-192

Solution for 0=-2n^2+40n-192 equation:

0=-2n^2+40n-192

We move all terms to the left:

0-(-2n^2+40n-192)=0

We add all the numbers together, and all the variables

-(-2n^2+40n-192)=0

We get rid of parentheses

2n^2-40n+192=0

a = 2; b = -40; c = +192;
Δ = b2-4ac
Δ = -402-4·2·192
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8}{2*2}=\frac{32}{4}
=8
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8}{2*2}=\frac{48}{4}
=12
$