(1+v)(5v+4)=0

Solution for (1+v)(5v+4)=0 equation:

(1+v)(5v+4)=0

We add all the numbers together, and all the variables

(v+1)(5v+4)=0

We multiply parentheses ..

(+5v^2+4v+5v+4)=0

We get rid of parentheses

5v^2+4v+5v+4=0

We add all the numbers together, and all the variables

5v^2+9v+4=0

a = 5; b = 9; c = +4;
Δ = b2-4ac
Δ = 92-4·5·4
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-1}{2*5}=\frac{-10}{10}
=-1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+1}{2*5}=\frac{-8}{10}
=-4/5
$