-2(v+3)(-v+5)=0

Solution for -2(v+3)(-v+5)=0 equation:

-2(v+3)(-v+5)=0

We add all the numbers together, and all the variables

-2(v+3)(-1v+5)=0

We multiply parentheses ..

-2(-1v^2+5v-3v+15)=0

We multiply parentheses

2v^2-10v+6v-30=0

We add all the numbers together, and all the variables

2v^2-4v-30=0

a = 2; b = -4; c = -30;
Δ = b2-4ac
Δ = -42-4·2·(-30)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-16}{2*2}=\frac{-12}{4}
=-3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+16}{2*2}=\frac{20}{4}
=5
$