v(6v+1)=0

Solution for v(6v+1)=0 equation:

v(6v+1)=0

We multiply parentheses

6v^2+v=0

a = 6; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·6·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*6}=\frac{-2}{12}
=-1/6
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*6}=\frac{0}{12}
=0
$