(3v+1)*(v+4)=0

Solution for (3v+1)*(v+4)=0 equation:

(3v+1)(v+4)=0

We multiply parentheses ..

(+3v^2+12v+v+4)=0

We get rid of parentheses

3v^2+12v+v+4=0

We add all the numbers together, and all the variables

3v^2+13v+4=0

a = 3; b = 13; c = +4;
Δ = b2-4ac
Δ = 132-4·3·4
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-11}{2*3}=\frac{-24}{6}
=-4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*3}=\frac{-2}{6}
=-1/3
$