2z^2+4z-6=0

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Solution for 2z^2+4z-6=0 equation: 



2z^2+4z-6=0

a = 2; b = 4; c = -6;
Δ = b2-4ac
Δ = 42-4·2·(-6)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*2}=\frac{-12}{4}
=-3
$


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*2}=\frac{4}{4}
=1
$

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