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

Solution for (5+v)(4v+2)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4v^2+2v+20v+10)=0

We get rid of parentheses

4v^2+2v+20v+10=0

We add all the numbers together, and all the variables

4v^2+22v+10=0

a = 4; b = 22; c = +10;
Δ = b2-4ac
Δ = 222-4·4·10
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-18}{2*4}=\frac{-40}{8}
=-5
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+18}{2*4}=\frac{-4}{8}
=-1/2
$