(5v+4)(3+v)=0

Solution for (5v+4)(3+v)=0 equation:

(5v+4)(3+v)=0

We add all the numbers together, and all the variables

(5v+4)(v+3)=0

We multiply parentheses ..

(+5v^2+15v+4v+12)=0

We get rid of parentheses

5v^2+15v+4v+12=0

We add all the numbers together, and all the variables

5v^2+19v+12=0

a = 5; b = 19; c = +12;
Δ = b2-4ac
Δ = 192-4·5·12
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{-(19)-11}{2*5}=\frac{-30}{10}
=-3
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*5}=\frac{-8}{10}
=-4/5
$