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

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

(V)=(20-2V)(20-2V)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+4V^2-40V-40V+400))+V=0


We calculate terms in parentheses: -((+4V^2-40V-40V+400)), so:

(+4V^2-40V-40V+400)

We get rid of parentheses

4V^2-40V-40V+400

We add all the numbers together, and all the variables

4V^2-80V+400

Back to the equation:

-(4V^2-80V+400)


We add all the numbers together, and all the variables

V-(4V^2-80V+400)=0

We get rid of parentheses

-4V^2+V+80V-400=0

We add all the numbers together, and all the variables

-4V^2+81V-400=0

a = -4; b = 81; c = -400;
Δ = b2-4ac
Δ = 812-4·(-4)·(-400)
Δ = 161
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{-(81)-\sqrt{161}}{2*-4}=\frac{-81-\sqrt{161}}{-8}
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(81)+\sqrt{161}}{2*-4}=\frac{-81+\sqrt{161}}{-8}
$