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

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

To find the range of the function for real values of $x$, let $y=f\left(x\right)$:
$y=\frac{2x-3}{2x^2+4x-6}$

Cross-multiply to clear the fraction and rearrange the equation into a quadratic form in terms of $x$:
$y(2x^2+4x-6)=2x-3$
$2y{x}^2+4yx-6y=2x-3$
$2y{x}^2+(4y-2)x+(3-6y)=0$

Since $x$ is a real number, the quadratic equation must have real roots. This requires the discriminant ($D$) to be greater than or equal to zero ($D\ge 0$):
$D=b^2-4ac\ge 0$
$(4y-2{)}^2-4(2y\left)\right(3-6y)\ge 0$

Expand and simplify the algebraic terms:
$(16y^2-16y+4)-8y(3-6y)\ge 0$
$16y^2-16y+4-24y+48y^2\ge 0$
$64y^2-40y+4\ge 0$

Divide the entire inequality by 4:
$16y^2-10y+1\ge 0$
Factor the quadratic inequality:
$(8y-1)(2y-1)\ge 0$

Solving $(8y-1)(2y-1)\ge 0$ yields:
$y\le \frac{1}{8}$ or $y\ge \frac{1}{2}$

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