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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+5v^2+20v+v+4)=0

We get rid of parentheses

5v^2+20v+v+4=0

We add all the numbers together, and all the variables

5v^2+21v+4=0

a = 5; b = 21; c = +4;
Δ = b2-4ac
Δ = 212-4·5·4
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-19}{2*5}=\frac{-40}{10}
=-4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+19}{2*5}=\frac{-2}{10}
=-1/5
$