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

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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 $

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