z^2+z-0.5=0

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Solution for z^2+z-0.5=0 equation: 



z^2+z-0.5=0

a = 1; b = 1; c = -0.5;
Δ = b2-4ac
Δ = 12-4·1·(-0.5)
Δ = 3
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


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

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