-4(v-2)=-4(v-8)8v

Solution for -4(v-2)=-4(v-8)8v equation:

-4(v-2)=-4(v-8)8v

We move all terms to the left:

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

We multiply parentheses

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


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

-4(v-8)8v

We multiply parentheses

-32v^2+256v

Back to the equation:

-(-32v^2+256v)


We get rid of parentheses

32v^2-256v-4v+8=0

We add all the numbers together, and all the variables

32v^2-260v+8=0

a = 32; b = -260; c = +8;
Δ = b2-4ac
Δ = -2602-4·32·8
Δ = 66576
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:

$\sqrt{\Delta}=\sqrt{66576}=\sqrt{16*4161}=\sqrt{16}*\sqrt{4161}=4\sqrt{4161}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-260)-4\sqrt{4161}}{2*32}=\frac{260-4\sqrt{4161}}{64}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-260)+4\sqrt{4161}}{2*32}=\frac{260+4\sqrt{4161}}{64}
$