(v+3)(v-15)=0

Solution for (v+3)(v-15)=0 equation:

(v+3)(v-15)=0

We multiply parentheses ..

(+v^2-15v+3v-45)=0

We get rid of parentheses

v^2-15v+3v-45=0

We add all the numbers together, and all the variables

v^2-12v-45=0

a = 1; b = -12; c = -45;
Δ = b2-4ac
Δ = -122-4·1·(-45)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-18}{2*1}=\frac{-6}{2}
=-3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+18}{2*1}=\frac{30}{2}
=15
$