2v^2+16v=0

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



2v^2+16v=0

a = 2; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·2·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$


$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*2}=\frac{-32}{4}
=-8
$


$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*2}=\frac{0}{4}
=0
$

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