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

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

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

We multiply parentheses ..

-(+5v^2-25v+3v-15)=0

We get rid of parentheses

-5v^2+25v-3v+15=0

We add all the numbers together, and all the variables

-5v^2+22v+15=0

a = -5; b = 22; c = +15;
Δ = b2-4ac
Δ = 222-4·(-5)·15
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-28}{2*-5}=\frac{-50}{-10}
=+5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+28}{2*-5}=\frac{6}{-10}
=-3/5
$