Question 945089
{{{(z^2-2)(z^3+i)=0}}}
You can both terms further,
{{{(z-sqrt(2))(z+sqrt(2))(z-i)(z^2+iz-1)=0}}}
So then, 
{{{z-sqrt(2)=0}}}
{{{highlight(z=sqrt(2))}}}
.
.
{{{z+sqrt(2)=0}}}
{{{highlight(z=-sqrt(2))}}}
.
.
{{{z-i=0}}}
{{{highlight(z=i)}}}
.
.
Complete the square for the final term,
{{{z^2+iz-1=0}}}
{{{z^2+iz+(i/2)^2-1=(i^2/2)}}}
{{{(z+i/2)^2=-1/4+1}}}
{{{(z+i/2)^2=3/4}}}
{{{z+i/2=0 +- sqrt(3)/2}}}
{{{highlight(z=-i/2 +- sqrt(3)/2)}}}