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

Solution for (5-v)(4v-1)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4v^2+v+20v-5)=0

We get rid of parentheses

-4v^2+v+20v-5=0

We add all the numbers together, and all the variables

-4v^2+21v-5=0

a = -4; b = 21; c = -5;
Δ = b2-4ac
Δ = 212-4·(-4)·(-5)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-19}{2*-4}=\frac{-40}{-8}
=+5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+19}{2*-4}=\frac{-2}{-8}
=1/4
$