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

The correct option is:
$$(-\infty ,\frac{1}{8}]\cup [\frac{1}{2},\infty )$$

Step-by-Step Explanation:

Let the given function be represented by y:
$$y=\frac{2x-3}{2x^2+4x-6}$$

First, we note that the domain of the function excludes values of x that make the denominator zero.
The denominator is:
$$2x^2+4x-6=2(x+3)(x-1)$$
Thus, the domain of the function is all real numbers x except $x=1$ and $x=-3$.

Now, we rewrite the equation by cross-multiplying:
$$y(2x^2+4x-6)=2x-3$$
Expanding and rearranging the terms to form a quadratic equation in terms of x:
$$2y{x}^2+4yx-6y=2x-3$$
$$2y{x}^2+(4y-2)x+(3-6y)=0$$

For x to be a real number, the discriminant D of this quadratic equation must be greater than or equal to zero (assuming $y\ne 0$).
The discriminant D is given by:
$$D=b^2-4ac\ge 0$$
Here, $a=2y$, $b=4y-2$, and $c=3-6y$. Substituting these values:
$${(4y-2)}^2-4\left(2y\right)(3-6y)\ge 0$$

Let's simplify the inequality step-by-step:
Factor out 2 from the term $(4y-2)$, which gives $2^2{(2y-1)}^2=4{(2y-1)}^2$.
Substitute this back:
$$4{(2y-1)}^2-8y(3-6y)\ge 0$$
Divide the entire inequality by 4:
$${(2y-1)}^2-2y(3-6y)\ge 0$$
Expand the terms:
$$(4y^2-4y+1)-6y+12y^2\ge 0$$
Combine like terms:
$$16y^2-10y+1\ge 0$$

Now, we factor the quadratic expression:
$$16y^2-8y-2y+1\ge 0$$
$$8y(2y-1)-1(2y-1)\ge 0$$
$$(8y-1)(2y-1)\ge 0$$

Solving this quadratic inequality yields:
$$y\le \frac{1}{8}\text{ or }y\ge \frac{1}{2}$$

Finally, we check if the boundary values are attainable by substituting them back into the quadratic equation in x:
If $y=\frac{1}{8}$, the equation becomes:
$$2\left(\frac{1}{8}\right)x^2+\left[4\right(\frac{1}{8})-2]x+[3-6(\frac{1}{8}\left)\right]=0$$
$$\frac{1}{4}{x}^2-\frac{3}{2}x+\frac{9}{4}=0\Rightarrow x^2-6x+9=0\Rightarrow {(x-3)}^2=0\Rightarrow x=3$$
Since x = 3 is a valid real number in the domain, $y=\frac{1}{8}$ is attainable.

If $y=\frac{1}{2}$, the equation becomes:
$$2\left(\frac{1}{2}\right)x^2+\left[4\right(\frac{1}{2})-2]x+[3-6(\frac{1}{2}\left)\right]=0$$
$$x^2+0x+0=0\Rightarrow x=0$$
Since x = 0 is a valid real number in the domain, $y=\frac{1}{2}$ is also attainable.

Thus, the range of the function is:
$$y\in (-\infty ,\frac{1}{8}]\cup [\frac{1}{2},\infty )$$