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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-5v^2-4v+35v+28)=0

We get rid of parentheses

-5v^2-4v+35v+28=0

We add all the numbers together, and all the variables

-5v^2+31v+28=0

a = -5; b = 31; c = +28;
Δ = b2-4ac
Δ = 312-4·(-5)·28
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-39}{2*-5}=\frac{-70}{-10}
=+7
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+39}{2*-5}=\frac{8}{-10}
=-4/5
$