V=(13-2x)(23-2x)

Solution for V=(13-2x)(23-2x) equation:

=(13-2V)(23-2V)

We move all terms to the left:

-((13-2V)(23-2V))=0

We add all the numbers together, and all the variables

-((-2V+13)(-2V+23))=0

We multiply parentheses ..

-((+4V^2-46V-26V+299))=0


We calculate terms in parentheses: -((+4V^2-46V-26V+299)), so:

(+4V^2-46V-26V+299)

We get rid of parentheses

4V^2-46V-26V+299

We add all the numbers together, and all the variables

4V^2-72V+299

Back to the equation:

-(4V^2-72V+299)


We get rid of parentheses

-4V^2+72V-299=0

a = -4; b = 72; c = -299;
Δ = b2-4ac
Δ = 722-4·(-4)·(-299)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-20}{2*-4}=\frac{-92}{-8}
=11+1/2
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+20}{2*-4}=\frac{-52}{-8}
=6+1/2
$