I'm fairly new to complex numbers and there wasn't much I could find online but could someone see if my answer is right? we need to find $z = a+bi$ given that $z^2 = -144+12i\sqrt{8}$
for LHS:
$$z^2 = (a+bi)^2 = a^2 + 2abi - b^2$$
$$a^2 +2abi - b^2 = -144+12i\sqrt{8}$$
equating real numbers : $$a^2 - b^2 = -144$$
equating imaginary numbers : $$2ab = 12\sqrt{8}$$
solving simultaneously gives
$$a = +1.4 \text{ or } -1.4 \\ b = +12.1 \text{ or } -12.1 $$
hence $z = 1.4 + 12.1i \text{ or } z = -1.4 - 12.1i$
Your calculation are correct, in fact $a,b$ are solutions to the following system: $$\left\{\begin{matrix} a^2-b^2=-144 \\ 2ab=12\sqrt{8} \end{matrix}\right.$$ From here, it's very easy to see that the solutions are: $$\left\{\begin{matrix} a=\frac{7}{5} \\ b=\frac{30\sqrt{8}}{7} \end{matrix}\right.$$ and: $$\left\{\begin{matrix} a=-\frac{7}{5} \\ b=-\frac{30\sqrt{8}}{7} \end{matrix}\right.$$ As a general rule, I don't advise you to trunc the solutions, but write: $$z_1=\frac{7}{5}+\frac{30\sqrt{8}}{7}i \: \vee \: z_2=-\frac{7}{5}-\frac{30\sqrt{8}}{7}i$$