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

Solution for (4v+5)(1-v)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4v^2-1v+5=0

a = -4; b = -1; c = +5;
Δ = b2-4ac
Δ = -12-4·(-4)·5
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-9}{2*-4}=\frac{-8}{-8}
=1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+9}{2*-4}=\frac{10}{-8}
=-1+1/4
$