V(x)=(16-2x)(11-2x)

Solution for V(x)=(16-2x)(11-2x) equation:

(V)=(16-2V)(11-2V)

We move all terms to the left:

(V)-((16-2V)(11-2V))=0

We add all the numbers together, and all the variables

V-((-2V+16)(-2V+11))=0

We multiply parentheses ..

-((+4V^2-22V-32V+176))+V=0


We calculate terms in parentheses: -((+4V^2-22V-32V+176)), so:

(+4V^2-22V-32V+176)

We get rid of parentheses

4V^2-22V-32V+176

We add all the numbers together, and all the variables

4V^2-54V+176

Back to the equation:

-(4V^2-54V+176)


We add all the numbers together, and all the variables

V-(4V^2-54V+176)=0

We get rid of parentheses

-4V^2+V+54V-176=0

We add all the numbers together, and all the variables

-4V^2+55V-176=0

a = -4; b = 55; c = -176;
Δ = b2-4ac
Δ = 552-4·(-4)·(-176)
Δ = 209
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{-(55)-\sqrt{209}}{2*-4}=\frac{-55-\sqrt{209}}{-8}
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(55)+\sqrt{209}}{2*-4}=\frac{-55+\sqrt{209}}{-8}
$