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

Solution for (v-1)(v+3)=0 equation:

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

We multiply parentheses ..

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

We get rid of parentheses

v^2+3v-1v-3=0

We add all the numbers together, and all the variables

v^2+2v-3=0

a = 1; b = 2; c = -3;
Δ = b2-4ac
Δ = 22-4·1·(-3)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4}{2*1}=\frac{-6}{2}
=-3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*1}=\frac{2}{2}
=1
$