(4v+1)(8-v)=v

Solution for (4v+1)(8-v)=v equation:

(4v+1)(8-v)=v

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4v^2+32v-1v+8)-1v=0

We get rid of parentheses

-4v^2+32v-1v-1v+8=0

We add all the numbers together, and all the variables

-4v^2+30v+8=0

a = -4; b = 30; c = +8;
Δ = b2-4ac
Δ = 302-4·(-4)·8
Δ = 1028
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:

$\sqrt{\Delta}=\sqrt{1028}=\sqrt{4*257}=\sqrt{4}*\sqrt{257}=2\sqrt{257}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{257}}{2*-4}=\frac{-30-2\sqrt{257}}{-8}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{257}}{2*-4}=\frac{-30+2\sqrt{257}}{-8}
$