(5v+2)(1-v)=0

Solution for (5v+2)(1-v)=0 equation:

(5v+2)(1-v)=0

We add all the numbers together, and all the variables

(5v+2)(-1v+1)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-5v^2+3v+2=0

a = -5; b = 3; c = +2;
Δ = b2-4ac
Δ = 32-4·(-5)·2
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*-5}=\frac{-10}{-10}
=1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*-5}=\frac{4}{-10}
=-2/5
$