-z+2z^2-5=0

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

-z+2z^2-5=0

We add all the numbers together, and all the variables

2z^2-1z-5=0

a = 2; b = -1; c = -5;
Δ = b2-4ac
Δ = -12-4·2·(-5)
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{-(-1)-\sqrt{41}}{2*2}=\frac{1-\sqrt{41}}{4}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{41}}{2*2}=\frac{1+\sqrt{41}}{4}
$