(5-v)(4v+7)=0

Solution for (5-v)(4v+7)=0 equation:

(5-v)(4v+7)=0

We add all the numbers together, and all the variables

(-1v+5)(4v+7)=0

We multiply parentheses ..

(-4v^2-7v+20v+35)=0

We get rid of parentheses

-4v^2-7v+20v+35=0

We add all the numbers together, and all the variables

-4v^2+13v+35=0

a = -4; b = 13; c = +35;
Δ = b2-4ac
Δ = 132-4·(-4)·35
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-27}{2*-4}=\frac{-40}{-8}
=+5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+27}{2*-4}=\frac{14}{-8}
=-1+3/4
$