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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4v^2+3v+20v-15)=0

We get rid of parentheses

-4v^2+3v+20v-15=0

We add all the numbers together, and all the variables

-4v^2+23v-15=0

a = -4; b = 23; c = -15;
Δ = b2-4ac
Δ = 232-4·(-4)·(-15)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-17}{2*-4}=\frac{-40}{-8}
=+5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+17}{2*-4}=\frac{-6}{-8}
=3/4
$