(3v-8)(4-v)=0

Solution for (3v-8)(4-v)=0 equation:

(3v-8)(4-v)=0

We add all the numbers together, and all the variables

(3v-8)(-1v+4)=0

We multiply parentheses ..

(-3v^2+12v+8v-32)=0

We get rid of parentheses

-3v^2+12v+8v-32=0

We add all the numbers together, and all the variables

-3v^2+20v-32=0

a = -3; b = 20; c = -32;
Δ = b2-4ac
Δ = 202-4·(-3)·(-32)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4}{2*-3}=\frac{-24}{-6}
=+4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4}{2*-3}=\frac{-16}{-6}
=2+2/3
$