3z2+3z-6=0

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Solution for 3z2+3z-6=0 equation: 



3z^2+3z-6=0

a = 3; b = 3; c = -6;
Δ = b2-4ac
Δ = 32-4·3·(-6)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$


$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*3}=\frac{-12}{6}
=-2
$


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*3}=\frac{6}{6}
=1
$

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