(4-v)(3v-5)=0

Solution for (4-v)(3v-5)=0 equation:

(4-v)(3v-5)=0

We add all the numbers together, and all the variables

(-1v+4)(3v-5)=0

We multiply parentheses ..

(-3v^2+5v+12v-20)=0

We get rid of parentheses

-3v^2+5v+12v-20=0

We add all the numbers together, and all the variables

-3v^2+17v-20=0

a = -3; b = 17; c = -20;
Δ = b2-4ac
Δ = 172-4·(-3)·(-20)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-7}{2*-3}=\frac{-24}{-6}
=+4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+7}{2*-3}=\frac{-10}{-6}
=1+2/3
$