4z^2+32z+60=0

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Solution for 4z^2+32z+60=0 equation: 



4z^2+32z+60=0

a = 4; b = 32; c = +60;
Δ = b2-4ac
Δ = 322-4·4·60
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{-(32)-8}{2*4}=\frac{-40}{8}
=-5
$


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8}{2*4}=\frac{-24}{8}
=-3
$

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