Question: top-left: 2.
center: By using a suitable substitution, solve the equation
center: $$(2x-3)^2 - \frac{4}{(2x-3)^2} - 3 = 0.$$
1. **State the problem:**
Solve the equation $$ (2x-3)^2 - \frac{4}{(2x-3)^2} - 3 = 0 $$ using a suitable substitution.
2. **Introduce substitution:**
Let $$ y = (2x-3)^2 $$.
The equation becomes:
$$ y - \frac{4}{y} - 3 = 0 $$.
3. **Multiply through by $$ y $$ to clear the denominator:**
$$ y \cdot y - y \cdot \frac{4}{y} - 3y = y^2 - 4 - 3y = 0 $$.
4. **Rewrite the quadratic equation:**
$$ y^2 - 3y - 4 = 0 $$.
5. **Factor the quadratic:**
$$ (y - 4)(y + 1) = 0 $$.
6. **Solve for $$ y $$:**
$$ y - 4 = 0 \implies y = 4 $$
$$ y + 1 = 0 \implies y = -1 $$
Since $$ y = (2x-3)^2 $$ and squares are always non-negative, discard $$ y = -1 $$.
7. **Solve for $$ x $$ when $$ y = 4 $$:**
$$ (2x - 3)^2 = 4 $$
Take square root on both sides:
$$ 2x - 3 = \pm 2 $$
8. **Find $$ x $$ values:**
For $$ 2x - 3 = 2 $$:
$$ 2x = 5 \implies x = \frac{5}{2} $$
For $$ 2x - 3 = -2 $$:
$$ 2x = 1 \implies x = \frac{1}{2} $$
**Final answer:**
$$ x = \frac{5}{2} \text{ or } x = \frac{1}{2} $$