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

Solution for (5-v)(4v+2)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4v^2-2v+20v+10)=0

We get rid of parentheses

-4v^2-2v+20v+10=0

We add all the numbers together, and all the variables

-4v^2+18v+10=0

a = -4; b = 18; c = +10;
Δ = b2-4ac
Δ = 182-4·(-4)·10
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-22}{2*-4}=\frac{-40}{-8}
=+5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*-4}=\frac{4}{-8}
=-1/2
$