z2+4z-5=0

Solution for z2+4z-5=0 equation:

z2+4z-5=0

We add all the numbers together, and all the variables

z^2+4z-5=0

a = 1; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·1·(-5)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6}{2*1}=\frac{-10}{2}
=-5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6}{2*1}=\frac{2}{2}
=1
$