(4v+2)(1+v)=0

Solution for (4v+2)(1+v)=0 equation:

(4v+2)(1+v)=0

We add all the numbers together, and all the variables

(4v+2)(v+1)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4v^2+6v+2=0

a = 4; b = 6; c = +2;
Δ = b2-4ac
Δ = 62-4·4·2
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2}{2*4}=\frac{-8}{8}
=-1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*4}=\frac{-4}{8}
=-1/2
$