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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+5v^2+4v+25v+20)=0

We get rid of parentheses

5v^2+4v+25v+20=0

We add all the numbers together, and all the variables

5v^2+29v+20=0

a = 5; b = 29; c = +20;
Δ = b2-4ac
Δ = 292-4·5·20
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-21}{2*5}=\frac{-50}{10}
=-5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+21}{2*5}=\frac{-8}{10}
=-4/5
$