2p^2+16p=0

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



2p^2+16p=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:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


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


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

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