2v^2+12v+16=0

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Solution for 2v^2+12v+16=0 equation: 



2v^2+12v+16=0

a = 2; b = 12; c = +16;
Δ = b2-4ac
Δ = 122-4·2·16
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4}{2*2}=\frac{-16}{4}
=-4
$


$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4}{2*2}=\frac{-8}{4}
=-2
$

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