3v(v+2)=1

Solution for 3v(v+2)=1 equation:

3v(v+2)=1

We move all terms to the left:

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

We multiply parentheses

3v^2+6v-1=0

a = 3; b = 6; c = -1;
Δ = b2-4ac
Δ = 62-4·3·(-1)
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{-(6)-4\sqrt{3}}{2*3}=\frac{-6-4\sqrt{3}}{6}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{3}}{2*3}=\frac{-6+4\sqrt{3}}{6}
$