Note: It's likely I'm off by some factors of 2 in places (or other similar mistakes). That's something the OP can check on their own.
There are a couple of issues. In the comments, OP mentions that their instructor says that they are solving the double-well potential. In that case, the Schrodinger equation reads
$$
-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}
- g(\delta(x-a)+\delta(x+a))\psi(x) = E\psi\,,
$$
which can be rearranged to get
$$
\left(\frac{\partial^2}{\partial x^2} +k^2\right)\psi
=
-\lambda(\delta(x-a)+\delta(x+a))\psi(x)
\,,
$$
where $k^2 = 2mE/\hbar^2$ and $\lambda = {2m}g/{\hbar^2}$. Right away, we see a sign issue: there should be a negative sign on the right-hand side if this is meant to be a double-well potential, in which case we should get two bound states.
Second, without going through the details, the Green's function for the Helmholtz equation, satisfying
$$
\left(\frac{\partial^2}{\partial x^2} +k^2\right)G_0(x) = -\delta(x)\,,
$$
is given by
$$
G_0(x) = i\frac{e^{ik|x|}}{2k}\,.
$$
Then, we multiply both sides of the rearranged Schrodinger equation by $G_0$ and integrate. For the left-hand side, we get
\begin{align}
\int dx\,G_0(x-x')\left(\frac{\partial^2}{\partial x^2} +k^2\right)\psi
=-\psi(x')\,,
\end{align}
after what amounts essentially to an integration by parts and applying the boundary conditions $\psi(x'\to\pm\infty)=0$ and $G_0(x'\to\pm\infty)=0$. On the right-hand side, we get
\begin{align*}
\int dx\,G_0(x-x')(-\lambda(\delta(x-a)+\delta(x+a))\psi(x))
&=
-\lambda\int dx\,(\delta(x-a)+\delta(x+a))i\frac{e^{-ik|x-x'|}}{2k}\psi(x)
\\
&=
-\lambda\left(i\frac{e^{ik|x'-a|}}{2k}\psi(a)+i\frac{e^{ik|x'+a|}}{2k}\psi(-a)\right)\,.
\end{align*}
So, we have the following relationship:
$$
\psi(x) = \lambda\left(i\frac{e^{ik|x-a|}}{2k}\psi(a)+i\frac{e^{ik|x+a|}}{2k}\psi(-a)\right)\,.
$$
Finally, how do we extract two linearly independent solution bound states from this? First of all, note that for the state to be bound, we need to have $E<0$, which implies that $k = \sqrt{2mE/\hbar^2} = i\kappa$, where $\kappa >0$. This guarantees that the resulting real exponentials decay at infinity. The second thing has to do with what happens at the positions of the delta functions. Evaluating the equation at $x=\pm a$, we get
\begin{align}
\psi(a) &= \lambda\left(\frac{e^{-\kappa|a-a|}}{2\kappa}\psi(a)+\frac{e^{-\kappa|a+a|}}{2\kappa}\psi(-a)\right)
=\frac{\lambda}{2\kappa}\left(\psi(a)+{e^{-2\kappa a}}\psi(-a)\right)\,,
\\
\psi(-a) &= \lambda\left(\frac{e^{-\kappa|-a-a|}}{2\kappa}\psi(a)+\frac{e^{-\kappa|-a+a|}}{2\kappa}\psi(-a)\right)
=\frac{\lambda}{2\kappa}\left({e^{-2\kappa}}\psi(a)+\psi(-a)\right)\,.
\end{align}
Solving the first for $\psi(-a)$, plugging it into the second, and rearranging yields
$$
\psi(a)
\left((\lambda -2 \kappa)e^{2 a \kappa }-\lambda \right)
\left((\lambda -2 \kappa )e^{2 a \kappa }+\lambda \right)=0\,,
$$
which is an implicit equation for $\kappa$. The first factor must be non-zero, and so we are left with the one of the second two factors being non-zero. One corresponds to the even bound state and the other corresponds the odd bound state.
