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

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

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

We multiply parentheses ..

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

We get rid of parentheses

-5v^2+5v-15v+15=0

We add all the numbers together, and all the variables

-5v^2-10v+15=0

a = -5; b = -10; c = +15;
Δ = b2-4ac
Δ = -102-4·(-5)·15
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-20}{2*-5}=\frac{-10}{-10}
=1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+20}{2*-5}=\frac{30}{-10}
=-3
$