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

Solution for -2z^2+2z+1=0 equation:

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

a = -2; b = 2; c = +1;
Δ = b2-4ac
Δ = 22-4·(-2)·1
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{3}}{2*-2}=\frac{-2-2\sqrt{3}}{-4}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{3}}{2*-2}=\frac{-2+2\sqrt{3}}{-4}
$