V(3)=(10-2x)(10-2x)

Solution for V(3)=(10-2x)(10-2x) equation:

(3)=(10-2V)(10-2V)

We move all terms to the left:

(3)-((10-2V)(10-2V))=0

We add all the numbers together, and all the variables

-((-2V+10)(-2V+10))+3=0

We multiply parentheses ..

-((+4V^2-20V-20V+100))+3=0


We calculate terms in parentheses: -((+4V^2-20V-20V+100)), so:

(+4V^2-20V-20V+100)

We get rid of parentheses

4V^2-20V-20V+100

We add all the numbers together, and all the variables

4V^2-40V+100

Back to the equation:

-(4V^2-40V+100)


We get rid of parentheses

-4V^2+40V-100+3=0

We add all the numbers together, and all the variables

-4V^2+40V-97=0

a = -4; b = 40; c = -97;
Δ = b2-4ac
Δ = 402-4·(-4)·(-97)
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{3}}{2*-4}=\frac{-40-4\sqrt{3}}{-8}
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{3}}{2*-4}=\frac{-40+4\sqrt{3}}{-8}
$