I think you made a mistake.
Using your wavefunction and noting that $$\int^\frac{L}{2}{-\frac{L}{2}} \sin\left(\frac{\pi x}{L}\right)\sin\left(\frac{2\pi x}{L}\right) = \frac{4L}{3\pi}$$ we see that $$1=\int^{L/2}{-L/2}|\psi(x)|^2 dx = \frac{2}{L}\left(|A_0|^2\frac{L}{2} +\frac{4L}{3\pi} \left(A_0 + A_0^\right)+ \frac{1}{4}\frac{L}{2}\right) = \left(|A_0|^2 +\frac{8}{3\pi} \left(A_0 + A_0^\right)+ \frac{1}{4}\right)$$ and so we have the constraint $$\boxed{|A_0|^2 +\frac{8}{3\pi} \left(A_0 + A_0^\right)+ \frac{1}{4} = 1}$$ If we let $A_0 = a+ib$, with $a,b$ real, this reduces to $$a^2+b^2 +\frac{16}{3\pi}a + \frac{1}{4} = 1 $$ If we make the choice that $A_0$ is real, that is $A_0 = a$ we find $$A_0^2 +\frac{16}{3\pi} A_0 + \frac{1}{4} = 1 ~~\implies ~~ \boxed{A_0 = -\frac{8}{3\pi}\pm\sqrt{\left(\frac{8}{3\pi}\right)^2+\frac{3}{4}}}$$ which is what I am guessing you were trying to get. We see that the most general solution involves both $a$ and $b$ - it is complex. To find a general result, parameterized by $a$ we solve $$a^2+b^2 +\frac{16}{3\pi}a + \frac{1}{4} = 1~~\implies ~~ b = \pm \sqrt{\frac{3}{4}-a^2-\frac{16}{3\pi}a}$$ and so our general solution is $$\boxed{A_0 = a \pm i\sqrt{\frac{3}{4}-a^2-\frac{16}{3\pi}a}} $$ (note we must restrict $a$ to take values for which $\sqrt{\frac{3}{4}-a^2-\frac{16}{3\pi}a}$ is real). So we have a range of possible solutions. 374
I cannot see any way to make a choice of one solution over any other.*
Added after comments below:
It is likely that the question (wherever you found it) has a typo and the interval of interest is actually $[0,L]$. This makes more sense, because the first two energy eigenstates of an infinite square well potential on the interval $[0,L]$ are (up to a normalization) $\sin\left(\frac{\pi x}{L}\right)$ and $\sin\left(\frac{2\pi x}{L}\right)$. Under this scenario, the problem of normalization is much simpler, because these two functions are orthogonal on this interval. This means that we have $$1 = \int^L_0 dx |\psi(x)|^2 = |A_0|^2 + \frac{1}{4} \implies |A_0| = \frac{\sqrt{3}}{2}$$ so we would have $A_0 = e^{i\theta}\frac{\sqrt{3}}{2}$ for arbitrary, real $\theta$. 474