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

Solution for (4v+3)(2+v)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4v^2+8v+3v+6)=0

We get rid of parentheses

4v^2+8v+3v+6=0

We add all the numbers together, and all the variables

4v^2+11v+6=0

a = 4; b = 11; c = +6;
Δ = b2-4ac
Δ = 112-4·4·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-5}{2*4}=\frac{-16}{8}
=-2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+5}{2*4}=\frac{-6}{8}
=-3/4
$