-.2+2i+i^2=0

Solution for -.2+2i+i^2=0 equation:

-.2+2i+i^2=0

We add all the numbers together, and all the variables

i^2+2i-0.2=0

a = 1; b = 2; c = -0.2;
Δ = b2-4ac
Δ = 22-4·1·(-0.2)
Δ = 4.8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-\sqrt{4.8}}{2*1}=\frac{-2-\sqrt{4.8}}{2}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{4.8}}{2*1}=\frac{-2+\sqrt{4.8}}{2}
$