z^2+5z-4=0

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Solution for z^2+5z-4=0 equation: 



z^2+5z-4=0

a = 1; b = 5; c = -4;
Δ = b2-4ac
Δ = 52-4·1·(-4)
Δ = 41
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{-(5)-\sqrt{41}}{2*1}=\frac{-5-\sqrt{41}}{2}
$


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{41}}{2*1}=\frac{-5+\sqrt{41}}{2}
$

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