z^2+0,8z+0,16=0

Solution for z^2+0,8z+0,16=0 equation:

z^2+0.8z+0.16=0

a = 1; b = 0.8; c = +0.16;
Δ = b2-4ac
Δ = 0.82-4·1·0.16
Δ = 1.1102230246252E-16
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{-(0.8)-\sqrt{1.1102230246252E-16}}{2*1}=\frac{-0.8-\sqrt{1.1102230246252E-16}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.8)+\sqrt{1.1102230246252E-16}}{2*1}=\frac{-0.8+\sqrt{1.1102230246252E-16}}{2}
$