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

Solution for (9-v)(4v+5)=0 equation:

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

We add all the numbers together, and all the variables

(-1v+9)(4v+5)=0

We multiply parentheses ..

(-4v^2-5v+36v+45)=0

We get rid of parentheses

-4v^2-5v+36v+45=0

We add all the numbers together, and all the variables

-4v^2+31v+45=0

a = -4; b = 31; c = +45;
Δ = b2-4ac
Δ = 312-4·(-4)·45
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-41}{2*-4}=\frac{-72}{-8}
=+9
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+41}{2*-4}=\frac{10}{-8}
=-1+1/4
$