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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4v^2+5v+16v-20)=0

We get rid of parentheses

-4v^2+5v+16v-20=0

We add all the numbers together, and all the variables

-4v^2+21v-20=0

a = -4; b = 21; c = -20;
Δ = b2-4ac
Δ = 212-4·(-4)·(-20)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-11}{2*-4}=\frac{-32}{-8}
=+4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+11}{2*-4}=\frac{-10}{-8}
=1+1/4
$