(5v-2)(v+4)=0

Solution for (5v-2)(v+4)=0 equation:

(5v-2)(v+4)=0

We multiply parentheses ..

(+5v^2+20v-2v-8)=0

We get rid of parentheses

5v^2+20v-2v-8=0

We add all the numbers together, and all the variables

5v^2+18v-8=0

a = 5; b = 18; c = -8;
Δ = b2-4ac
Δ = 182-4·5·(-8)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-22}{2*5}=\frac{-40}{10}
=-4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*5}=\frac{4}{10}
=2/5
$