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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4v^2+20v-1v+5)=0

We get rid of parentheses

-4v^2+20v-1v+5=0

We add all the numbers together, and all the variables

-4v^2+19v+5=0

a = -4; b = 19; c = +5;
Δ = b2-4ac
Δ = 192-4·(-4)·5
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*-4}=\frac{-40}{-8}
=+5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*-4}=\frac{2}{-8}
=-1/4
$