4z^2+8z-12=0

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Solution for 4z^2+8z-12=0 equation: 



4z^2+8z-12=0

a = 4; b = 8; c = -12;
Δ = b2-4ac
Δ = 82-4·4·(-12)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$


$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*4}=\frac{-24}{8}
=-3
$


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*4}=\frac{8}{8}
=1
$

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