6v^2+2v=1

Solution for 6v^2+2v=1 equation:

6v^2+2v=1

We move all terms to the left:

6v^2+2v-(1)=0

a = 6; b = 2; c = -1;
Δ = b2-4ac
Δ = 22-4·6·(-1)
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{7}}{2*6}=\frac{-2-2\sqrt{7}}{12}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{7}}{2*6}=\frac{-2+2\sqrt{7}}{12}
$