Help solving the quadratic equation $ax^2-4bx+4bc-\frac{d^2}{a}=0$

Question

I have been struggling to solve this quadratic equation in the variable $x$ with integral coefficients: $$ax^2-4bx+4bc-\frac{d^2}{a}=0$$ $a\neq 0$ of course.How do I ensure that $x$ is an integer?

What I have done:

$$\Delta^2=(-4b)^2-4a(4bc-\frac{d^2}{a})$$

$$\Delta^2=16b^2-16abc+4d^2$$

I know that $d^2\equiv 0\pmod a$. So there exists a non-zero $k$ such that $d^2=ak$

$$\Delta^2=16b^2-16abc+4(ak)^2$$

$$(\dfrac{\Delta}{2})^2=(2b)^2-2(2b)ac+(ak)^2$$ Can I conclude that $c=\pm k$? How do I proceed from here?