(4-v)(3v+2)=0

Solution for (4-v)(3v+2)=0 equation:

(4-v)(3v+2)=0

We add all the numbers together, and all the variables

(-1v+4)(3v+2)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3v^2+10v+8=0

a = -3; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·(-3)·8
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-14}{2*-3}=\frac{-24}{-6}
=+4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14}{2*-3}=\frac{4}{-6}
=-2/3
$