3(2v^2-4v+1)=0

Solution for 3(2v^2-4v+1)=0 equation:

3(2v^2-4v+1)=0

We multiply parentheses

6v^2-12v+3=0

a = 6; b = -12; c = +3;
Δ = b2-4ac
Δ = -122-4·6·3
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{2}}{2*6}=\frac{12-6\sqrt{2}}{12}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{2}}{2*6}=\frac{12+6\sqrt{2}}{12}
$