-4=-8v^2+v

Solution for -4=-8v^2+v equation:

-4=-8v^2+v

We move all terms to the left:

-4-(-8v^2+v)=0

We get rid of parentheses

8v^2-v-4=0

We add all the numbers together, and all the variables

8v^2-1v-4=0

a = 8; b = -1; c = -4;
Δ = b2-4ac
Δ = -12-4·8·(-4)
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{129}}{2*8}=\frac{1-\sqrt{129}}{16}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{129}}{2*8}=\frac{1+\sqrt{129}}{16}
$