V(x)=(21-2x)(16-2x)

Solution for V(x)=(21-2x)(16-2x) equation:

(V)=(21-2V)(16-2V)

We move all terms to the left:

(V)-((21-2V)(16-2V))=0

We add all the numbers together, and all the variables

V-((-2V+21)(-2V+16))=0

We multiply parentheses ..

-((+4V^2-32V-42V+336))+V=0


We calculate terms in parentheses: -((+4V^2-32V-42V+336)), so:

(+4V^2-32V-42V+336)

We get rid of parentheses

4V^2-32V-42V+336

We add all the numbers together, and all the variables

4V^2-74V+336

Back to the equation:

-(4V^2-74V+336)


We add all the numbers together, and all the variables

V-(4V^2-74V+336)=0

We get rid of parentheses

-4V^2+V+74V-336=0

We add all the numbers together, and all the variables

-4V^2+75V-336=0

a = -4; b = 75; c = -336;
Δ = b2-4ac
Δ = 752-4·(-4)·(-336)
Δ = 249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(75)-\sqrt{249}}{2*-4}=\frac{-75-\sqrt{249}}{-8}
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(75)+\sqrt{249}}{2*-4}=\frac{-75+\sqrt{249}}{-8}
$