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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(8v-4v(v+8)), so:

8v-4v(v+8)

We multiply parentheses

-4v^2+8v-32v

We add all the numbers together, and all the variables

-4v^2-24v

Back to the equation:

-(-4v^2-24v)


We get rid of parentheses

4v^2+24v+8=0

a = 4; b = 24; c = +8;
Δ = b2-4ac
Δ = 242-4·4·8
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{7}}{2*4}=\frac{-24-8\sqrt{7}}{8}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{7}}{2*4}=\frac{-24+8\sqrt{7}}{8}
$