V=(32-2x)*(20-2x)*2

Solution for V=(32-2x)*(20-2x)*2 equation:

=(32-2V)(20-2V)*2

We move all terms to the left:

-((32-2V)(20-2V)*2)=0

We add all the numbers together, and all the variables

-((-2V+32)(-2V+20)*2)=0

We multiply parentheses ..

-((+4V^2-40V-64V+640)*2)=0


We calculate terms in parentheses: -((+4V^2-40V-64V+640)*2), so:

(+4V^2-40V-64V+640)*2

We multiply parentheses

8V^2-80V-128V+1280

We add all the numbers together, and all the variables

8V^2-208V+1280

Back to the equation:

-(8V^2-208V+1280)


We get rid of parentheses

-8V^2+208V-1280=0

a = -8; b = 208; c = -1280;
Δ = b2-4ac
Δ = 2082-4·(-8)·(-1280)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(208)-48}{2*-8}=\frac{-256}{-16}
=+16
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(208)+48}{2*-8}=\frac{-160}{-16}
=+10
$