2v^2+6v+4=0

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Solution for 2v^2+6v+4=0 equation: 



2v^2+6v+4=0

a = 2; b = 6; c = +4;
Δ = b2-4ac
Δ = 62-4·2·4
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2}{2*2}=\frac{-8}{4}
=-2
$


$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*2}=\frac{-4}{4}
=-1
$

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