(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+17v+15=0

a = -4; b = 17; c = +15;
Δ = b2-4ac
Δ = 172-4·(-4)·15
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-23}{2*-4}=\frac{-40}{-8}
=+5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+23}{2*-4}=\frac{6}{-8}
=-3/4
$