v^2+4v=-1+8v

Solution for v^2+4v=-1+8v equation:

v^2+4v=-1+8v

We move all terms to the left:

v^2+4v-(-1+8v)=0

We add all the numbers together, and all the variables

v^2+4v-(8v-1)=0

We get rid of parentheses

v^2+4v-8v+1=0

We add all the numbers together, and all the variables

v^2-4v+1=0

a = 1; b = -4; c = +1;
Δ = b2-4ac
Δ = -42-4·1·1
Δ = 12
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:

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