For real values of x, the range of the function $$f(x)=\dfrac{2x-3}{2x^{2}+4x-6}$$ is

Let $$y= \dfrac{2x-3}{2x^{2}+4x-6}$$

$$\Rightarrow 2yx^2 + 4yx - 6y = 2x - 3$$

$$\Rightarrow 2yx^2 + (4y-2)x - 6y + 3 = 0$$ 

Equation (1) is a quadratic in $$x$$, where $$x$$ is real. Therefore, the discriminant of the quadratic has to be greater than or equal to zero.

$$(4y-2)^2 + 4\times 2y\times (6y-3) \geq 0$$

$$16y^2 + 4 - 16y - 24y + 48y^2 \geq 0$$

$$64y^2 - 40y + 4 \geq 0$$

$$16y^2 - 10y + 1 \geq 0$$ 

The roots of the quadratic above will be $$\dfrac{10\pm 6}{32} = \dfrac{1}{2}$$ or $$\dfrac{1}{8}$$

Since the coefficient of $$y^2$$ is positive, the quadratic will be less than zero in the range $$\left(\dfrac{1}{8},\dfrac{1}{2}\right)$$

The quadratic will be greater than or equal to zero otherwise.

Therefore, the domain the quadratic, or possible values of $$y$$, which is the range of $$f(x)$$, will be,

$$\left(-\infty, \dfrac{1}{8}\right] \cup \left[\dfrac{1}{2}, \infty \right)$$

Option C is the correct answer.