zz*+3(z-z*) = 13+12i

Substitute x+yi for z and x-yi for conjugate z*,
where x and y are real.

(x+yi)(x-yi)+3[(x+yi)-(x-yi)] = 13+12i

Simplify the left side:

x²-y²i²+3[x+yi-x+yi] = 13+12i

Simplify using the fact that i² = -1

x²-y²(-1)+3[2yi] = 13+12i

x²+y²+6yi = 13+12i

Set the sum of the two real-numbered terms on the 
left side, x²+y² equal to the real term, 13, on 
the right side:

x²+y² = 13  <--equation #1

Set the sum of the two imaginary term on the 
left side, 6yi equal to the imaginary term, 12i, on 
the right side:

6yi = 12i

Divide both sides by 6i

y = 2

Substitute 2 for y in 

x²+y² = 13

x²+2² = 13
 x²+4 = 13
   x² = 9
    x = ±3

We get two answers: 

z = x+yi = ±3+2i

So the two answers are z = 3+2i and x = -3+2i 

Edwin