-z2+10z-2=0

Solution for -z2+10z-2=0 equation:

-z2+10z-2=0

We add all the numbers together, and all the variables

-1z^2+10z-2=0

a = -1; b = 10; c = -2;
Δ = b2-4ac
Δ = 102-4·(-1)·(-2)
Δ = 92
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:

$\sqrt{\Delta}=\sqrt{92}=\sqrt{4*23}=\sqrt{4}*\sqrt{23}=2\sqrt{23}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{23}}{2*-1}=\frac{-10-2\sqrt{23}}{-2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{23}}{2*-1}=\frac{-10+2\sqrt{23}}{-2}
$