i^2+2i-1=0

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

i^2+2i-1=0

a = 1; b = 2; c = -1;
Δ = b2-4ac
Δ = 22-4·1·(-1)
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{2}}{2*1}=\frac{-2-2\sqrt{2}}{2}
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{2}}{2*1}=\frac{-2+2\sqrt{2}}{2}
$