(3z^2)+z=0

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Solution for (3z^2)+z=0 equation: 



(3z^2)+z=0

a = 3; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·3·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$


$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*3}=\frac{-2}{6}
=-1/3
$


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*3}=\frac{0}{6}
=0
$

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