First, let's expand out everything. I'm not going to show the steps, but we eventually get
\(2z^4-32z^3+204z^2-608z+706=-8\\2z^4-32z^3+204z^2-608z+714=0\)
Now, since every number is divisible by 2, we get factor out 2 to get
\({2\left(z^{4}-16z^{3}+102z^{2}-304z+357\right)}=0\)
Now, note that \(z^{4}-16z^{3}+102z^{2}-304z+357=(z^{2}-8z+17)(z^{2}-8z+21)\)
This means that we have the equation
\(2\left(z^{2}-8z+17\right)\left(z^{2}-8z+21\right)=0\)
Since the equation is equal to 0, we get two equations. We have
\(z^{2}-8z+17=0,\qquad z^{2}-8z+21=0\)
Using the quadratic equation on both formulas, we find 4 values of z.
We have
\(z=4+i\\ z=4-i\\ z=4+\sqrt{5}\cdot (i)\\ z=4+\sqrt{5}\cdot (-i)\)
This was a very basic explenation of the problem.
I can go in dephth more if you want to.
Thanks! :)