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

Solution for -4(v-2)(-v+5)=0 equation:

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

We add all the numbers together, and all the variables

-4(v-2)(-1v+5)=0

We multiply parentheses ..

-4(-1v^2+5v+2v-10)=0

We multiply parentheses

4v^2-20v-8v+40=0

We add all the numbers together, and all the variables

4v^2-28v+40=0

a = 4; b = -28; c = +40;
Δ = b2-4ac
Δ = -282-4·4·40
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-12}{2*4}=\frac{16}{8}
=2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+12}{2*4}=\frac{40}{8}
=5
$