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

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

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

We move all terms to the left:

8-(-8v-4v(v+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^2-40v

Back to the equation:

-(-4v^2-40v)


We get rid of parentheses

4v^2+40v+8=0

a = 4; b = 40; c = +8;
Δ = b2-4ac
Δ = 402-4·4·8
Δ = 1472
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:

$\sqrt{\Delta}=\sqrt{1472}=\sqrt{64*23}=\sqrt{64}*\sqrt{23}=8\sqrt{23}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{23}}{2*4}=\frac{-40-8\sqrt{23}}{8}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{23}}{2*4}=\frac{-40+8\sqrt{23}}{8}
$