⚡ AZUCATION SHORTCUT:
Let $y = f(x)$. Form a quadratic in $x$: $2yx^2 + (4y-2)x + (3-6y) = 0$.
For $x$ to be real, Discriminant $D \ge 0$.
$(4y-2)^2 - 4(2y)(3-6y) \ge 0 \implies 64y^2 - 40y + 4 \ge 0 \implies 16y^2 - 10y + 1 \ge 0$.
Roots are $1/2$ and $1/8$. Since it's $\ge 0$, $y \in (-\infty, 1/8] \cup [1/2, \infty)$.

Step 1: Set the function equal to $y$
$y = \dfrac{2x-3}{2x^2+4x-6}$
Cross-multiplying gives:
$y(2x^2 + 4x - 6) = 2x - 3$
$2yx^2 + 4yx - 6y - 2x + 3 = 0$
$2yx^2 + (4y - 2)x + (3 - 6y) = 0$

Step 2: Condition for real $x$
For $x$ to be real, the discriminant of this quadratic equation must be greater than or equal to zero ($D \ge 0$):
$B^2 - 4AC \ge 0$
$(4y - 2)^2 - 4(2y)(3 - 6y) \ge 0$
$16y^2 - 16y + 4 - 8y(3 - 6y) \ge 0$
$16y^2 - 16y + 4 - 24y + 48y^2 \ge 0$
$64y^2 - 40y + 4 \ge 0$

Step 3: Solve the inequality
Divide the entire inequality by 4:
$16y^2 - 10y + 1 \ge 0$
Factorizing the quadratic: $(8y - 1)(2y - 1) \ge 0$
The critical points are $y = \frac{1}{8}$ and $y = \frac{1}{2}$.
For the expression to be $\ge 0$, $y$ must lie outside the roots:
$y \in (-\infty, \frac{1}{8}] \cup [\frac{1}{2}, \infty)$.

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