2y(y-4)+3x(ax+4)=xy(2a+3) equation

Move right part of the equation to
left part with negative sign.

The equation is transformed from
$$3 x \left(a x + 4\right) + 2 y \left(y - 4\right) = x y \left(2 a + 3\right)$$
to
$$- x y \left(2 a + 3\right) + \left(3 x \left(a x + 4\right) + 2 y \left(y - 4\right)\right) = 0$$
Expand the expression in the equation
$$- x y \left(2 a + 3\right) + \left(3 x \left(a x + 4\right) + 2 y \left(y - 4\right)\right) = 0$$
We get the quadratic equation
$$3 a x^{2} - 2 a x y - 3 x y + 2 y^{2} + 12 x - 8 y = 0$$
This equation is of the form
$$a\ x^2 + b\ x + c = 0$$
A quadratic equation can be solved using the discriminant
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 3 a$$
$$b = - 2 a y - 3 y + 12$$
$$c = 2 y^{2} - 8 y$$
, then
$$D = b^2 - 4\ a\ c = $$
$$- 4 \cdot 3 a \left(2 y^{2} - 8 y\right) + \left(- 2 a y - 3 y + 12\right)^{2} = - 12 a \left(2 y^{2} - 8 y\right) + \left(- 2 a y - 3 y + 12\right)^{2}$$
The equation has two roots.
$$x_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$x_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$x_{1} = \frac{2 a y + 3 y + \sqrt{- 12 a \left(2 y^{2} - 8 y\right) + \left(- 2 a y - 3 y + 12\right)^{2}} - 12}{6 a}$$
Simplify
$$x_{2} = \frac{2 a y + 3 y - \sqrt{- 12 a \left(2 y^{2} - 8 y\right) + \left(- 2 a y - 3 y + 12\right)^{2}} - 12}{6 a}$$
Simplify