1. Problem statement. Problem: Solve the inequality $$ (2 - 5x)^2 > 4 (x - 3)^2 $$. 2. Formula and key rule. We use the identity that for real numbers $a,b$ we have $a^2 > b^2 \iff |a| > |b|$. Applying this gives $$|2-5x| > 2|x-3|$$. The important rule is to split into cases using the zeros of the linear expressions inside absolute values. 3. Critical points and cases. The critical points are $x = \frac{2}{5}$ and $x = 3$. We solve the inequality on the intervals determined by these points. 4. Case 1: $x < \frac{2}{5}$. Here $2-5x>0$ and $x-3<0$, so $|2-5x|=2-5x$ and $|x-3|=3-x$. Solve $$2-5x > 2(3-x)$$. Compute: $$2-5x > 6-2x$$. Bring like terms together: $$-3x > 4$$. Divide both sides by $-3$ (flip inequality): $$\frac{\cancel{-3}x}{\cancel{-3}} < \frac{4}{-3}$$. Thus $x < -\frac{4}{3}$. Intersecting with the case condition $x < \frac{2}{5}$ gives the solution part $x < -\frac{4}{3}$. 5. Case 2: $\frac{2}{5} \le x < 3$. Here $2-5x\le 0$ and $x-3<0$, so $|2-5x|=5x-2$ and $|x-3|=3-x$. Solve $$5x-2 > 2(3-x)$$. Compute: $$5x-2 > 6-2x$$. Bring like terms together: $$7x > 8$$. Divide both sides by $7$: $$\frac{\cancel{7}x}{\cancel{7}} > \frac{8}{7}$$. Thus $x > \frac{8}{7}$. Intersecting with the case condition $\frac{2}{5} \le x < 3$ gives the solution part $\frac{8}{7} < x < 3$. 6. Case 3: $x \ge 3$. Here $2-5x\le 0$ and $x-3\ge 0$, so $|2-5x|=5x-2$ and $|x-3|=x-3$. Solve $$5x-2 > 2(x-3)$$. Compute: $$5x-2 > 2x-6$$. Bring like terms together: $$3x > -4$$. Divide both sides by $3$: $$\frac{\cancel{3}x}{\cancel{3}} > \frac{-4}{3}$$. Thus $x > -\frac{4}{3}$. Intersecting with the case condition $x \ge 3$ gives the solution part $x \ge 3$. 7. Combine the solution parts from all cases. From Case 1 we have $x < -\frac{4}{3}$. From Case 2 and Case 3 combined we get $x > \frac{8}{7}$, which with Case 3 extends to $x \ge 3$ and thus to all $x > \frac{8}{7}$. Therefore the full solution set is $x \in (-\infty, -\frac{4}{3}) \cup (\frac{8}{7}, \infty)$. Final answer: $x \in (-\infty, -\frac{4}{3}) \cup (\frac{8}{7}, \infty)$.