20p^2+6p=0

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Solution for 20p^2+6p=0 equation: 



20p^2+6p=0

a = 20; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·20·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*20}=\frac{-12}{40}
=-3/10
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*20}=\frac{0}{40}
=0
$

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