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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-2v^2+8v-1v+4)=0

We get rid of parentheses

-2v^2+8v-1v+4=0

We add all the numbers together, and all the variables

-2v^2+7v+4=0

a = -2; b = 7; c = +4;
Δ = b2-4ac
Δ = 72-4·(-2)·4
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-9}{2*-2}=\frac{-16}{-4}
=+4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+9}{2*-2}=\frac{2}{-4}
=-1/2
$