-6z^2+3z=0

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Solution for -6z^2+3z=0 equation: 



-6z^2+3z=0

a = -6; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-6)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$


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


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-6}=\frac{0}{-12}
=0
$

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