Question 1098569
Let {{{z=a+bi}}}, so then
{{{z^2=(a+bi)(a+bi)=a^2-b^2+(2ab)i}}}
So then in this case,
1.{{{a^2-b^2=-5}}}
2.{{{2ab=12}}}
From 2,
{{{ab=6}}}
{{{a=6/b}}}
{{{a^2=36/b^2}}}
Substituting into 1,
{{{36/b^2-b^2=-5}}}
{{{36-b^4=-5b^2}}}
{{{b^4-5b^2-36=0}}}
Use a substitution,
{{{u=b^2}}}
{{{u^2-5u-36=0}}}
{{{(u-9)(u+4)=0}}}
So there are two u solutions,
{{{u-9=0}}}
{{{u=9}}}
Since we assume that {{{a}}} and {{{b}}} are real numbers we only solve for this solution. 
The other would lead to an imaginary number.
So then when {{{u=9}}},
{{{b^2=9}}}
{{{b=0 +- 3}}}
When {{{b=0 +-3 }}},
{{{a(0 +- 3)=6}}}
{{{a=0 +- 2}}},
From eq. 2, the product is positive so a and b have the same sign.
{{{z=2+3i)}}} and {{{z=-(2+3i)}}}