Solve in terms of a+bi=z ... z = (5−3i )(−2−3i +z)

Question says it all

z = (5-3i)(-2-3i+z) = -10-15i+5z+6i+9i^2-3iz = -19-9i+5z-3iz = (-19-9i) + (5-3i)z
z-(5-3i)z = (-19-9i)
z(1-5+3i) = (-19-9i)
z = (-19-9i)/(-4+3i) = (-19-9i)(-4-3i) / ((-4+3i)(-4-3i)) = 76+57i+36i-27/(16+9) = (49+93i)/25
z = 49/25+(93/25)i

Substitute a+bi into the equation in place of z.
Then multiply out which gives a+bi=(-19+5a+3b)+(-9-3a+5b)i
So a=-19+5a+3b and b=-9-3a+5b
Solve simultaneously to give a=(49/25) and b=(93/25)
Therefore z=(49/25)+(93/25)i.