6v^2=1+v

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

6v^2=1+v

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

6v^2-v-1=0

We add all the numbers together, and all the variables

6v^2-1v-1=0

a = 6; b = -1; c = -1;
Δ = b2-4ac
Δ = -12-4·6·(-1)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*6}=\frac{-4}{12}
=-1/3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*6}=\frac{6}{12}
=1/2
$