0=(v+1)(v+5)

Solution for 0=(v+1)(v+5) equation:

0=(v+1)(v+5)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+v^2+5v+v+5)), so:

(+v^2+5v+v+5)

We get rid of parentheses

v^2+5v+v+5

We add all the numbers together, and all the variables

v^2+6v+5

Back to the equation:

-(v^2+6v+5)


We get rid of parentheses

-v^2-6v-5=0

We add all the numbers together, and all the variables

-1v^2-6v-5=0

a = -1; b = -6; c = -5;
Δ = b2-4ac
Δ = -62-4·(-1)·(-5)
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{-(-6)-4}{2*-1}=\frac{2}{-2}
=-1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4}{2*-1}=\frac{10}{-2}
=-5
$