V(x)=(11-2x)(8-2x)

Solution for V(x)=(11-2x)(8-2x) equation:

(V)=(11-2V)(8-2V)

We move all terms to the left:

(V)-((11-2V)(8-2V))=0

We add all the numbers together, and all the variables

V-((-2V+11)(-2V+8))=0

We multiply parentheses ..

-((+4V^2-16V-22V+88))+V=0


We calculate terms in parentheses: -((+4V^2-16V-22V+88)), so:

(+4V^2-16V-22V+88)

We get rid of parentheses

4V^2-16V-22V+88

We add all the numbers together, and all the variables

4V^2-38V+88

Back to the equation:

-(4V^2-38V+88)


We add all the numbers together, and all the variables

V-(4V^2-38V+88)=0

We get rid of parentheses

-4V^2+V+38V-88=0

We add all the numbers together, and all the variables

-4V^2+39V-88=0

a = -4; b = 39; c = -88;
Δ = b2-4ac
Δ = 392-4·(-4)·(-88)
Δ = 113
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{-(39)-\sqrt{113}}{2*-4}=\frac{-39-\sqrt{113}}{-8}
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+\sqrt{113}}{2*-4}=\frac{-39+\sqrt{113}}{-8}
$