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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4v^2+9v+2=0

a = 4; b = 9; c = +2;
Δ = b2-4ac
Δ = 92-4·4·2
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-7}{2*4}=\frac{-16}{8}
=-2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+7}{2*4}=\frac{-2}{8}
=-1/4
$