3s^2+27s=0

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Solution for 3s^2+27s=0 equation: 



3s^2+27s=0

a = 3; b = 27; c = 0;
Δ = b2-4ac
Δ = 272-4·3·0
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$


$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-27}{2*3}=\frac{-54}{6}
=-9
$


$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+27}{2*3}=\frac{0}{6}
=0
$

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