0=-5z^2-4z+5

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

0=-5z^2-4z+5

We move all terms to the left:

0-(-5z^2-4z+5)=0

We add all the numbers together, and all the variables

-(-5z^2-4z+5)=0

We get rid of parentheses

5z^2+4z-5=0

a = 5; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·5·(-5)
Δ = 116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{29}}{2*5}=\frac{-4-2\sqrt{29}}{10}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{29}}{2*5}=\frac{-4+2\sqrt{29}}{10}
$