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

The denominator can be factored as:

$2x^2+4x-6 = 2(x+3)(x-1)$

The function is undefined at $x = 1$ and $x = -3$.

---

Let $y = f(x)$ and cross-multiply:

$y(2x^2+4x-6) = 2x-3$

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

For $x$ to be real, the discriminant of this quadratic must be $\geq 0$.

---

When $y = 0$, the equation becomes:

$-2x + 3 = 0$

$x = \dfrac{3}{2}$

This is a valid $x$-value (not excluded from the domain), so $y = 0$ is in the range.

---

When $y \neq 0$, applying the discriminant condition $\Delta \geq 0$:

$\Delta = (4y-2)^2 - 4(2y)(3-6y) \geq 0$

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

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

$= 4(16y^2 - 10y + 1) \geq 0$

$16y^2 - 10y + 1 \geq 0$

---

Roots of $16y^2 - 10y + 1 = 0$:

$y = \dfrac{10 \pm \sqrt{100 - 64}}{32}$

$= \dfrac{10 \pm 6}{32}$

$y = \dfrac{1}{8}$ or $y = \dfrac{1}{2}$

Since the coefficient of $y^2$ is positive, the parabola opens upward, so the expression is $\geq 0$ outside the roots:

$y \leq \dfrac{1}{8}$ or $y \geq \dfrac{1}{2}$

---

Checking whether the excluded points $x = 1$ and $x = -3$ affect the range:

Substituting $x = 1$ into $2yx^2 + (4y-2)x + (3-6y) = 0$ gives $1 = 0$.

Substituting $x = -3$ gives $9 = 0$.

Neither excluded point is ever a solution, so no $y$-values need to be removed.

---

Range of $f(x) = \left(-\infty,\, \dfrac{1}{8}\right] \cup \left[\dfrac{1}{2},\, \infty\right)$

The correct answer is Option 2.