6v^2+16v+8=0

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Solution for 6v^2+16v+8=0 equation: 



6v^2+16v+8=0

a = 6; b = 16; c = +8;
Δ = b2-4ac
Δ = 162-4·6·8
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*6}=\frac{-24}{12}
=-2
$


$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*6}=\frac{-8}{12}
=-2/3
$

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