z^2+2(1-I)z+7I=0

Solution for z^2+2(1-I)z+7I=0 equation:

z^2+2(1-)z+7=0

We add all the numbers together, and all the variables

z^2+20z+7=0

a = 1; b = 20; c = +7;
Δ = b2-4ac
Δ = 202-4·1·7
Δ = 372
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:

$\sqrt{\Delta}=\sqrt{372}=\sqrt{4*93}=\sqrt{4}*\sqrt{93}=2\sqrt{93}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{93}}{2*1}=\frac{-20-2\sqrt{93}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{93}}{2*1}=\frac{-20+2\sqrt{93}}{2}
$