V(2)=4x^2-2x+1

Solution for V(2)=4x^2-2x+1 equation:

(2)=4V^2-2V+1

We move all terms to the left:

(2)-(4V^2-2V+1)=0

We get rid of parentheses

-4V^2+2V-1+2=0

We add all the numbers together, and all the variables

-4V^2+2V+1=0

a = -4; b = 2; c = +1;
Δ = b2-4ac
Δ = 22-4·(-4)·1
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{5}}{2*-4}=\frac{-2-2\sqrt{5}}{-8}
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{5}}{2*-4}=\frac{-2+2\sqrt{5}}{-8}
$