So it is pretty straight forward, solve this.
The Attempt at a Solution
So what I would do first is solve 2(2-1)z, I get (2-2i)z=2z-2iz
we now have z2-2iz+7i+2z=0
Now I don't really know what to do because my textbook has two examples, in both z2 is ignored.
first it has z2+2iz-1-i=0 and used a=1, b=2i and c=-1-i
the second example shows z2+2z+4=0 and has 2z=b and ac =4*1
the problem with the two examples is I cannot deduce what will be a, b, and c in my problem.
I mean following the logic of the first example I get 2z = b or -2iz = b and then c = either 7i or 7i + 2z or something completely different.
I tried plugging in the numbers
(-2i ±√(2i^2-4*7i))/2 = (-2i±√(-4-28i))/2
then I tried 2z = b instead of 2i = b
(-2 ±√(2^2-4*7i))/2 = (-2±√(4-28i))/2
I mean this is just guessing and I kept going, what if 7i = b, etc. but it doesnt help me understand what it should be and why, which is really what I want to know and not the solution.