1. **State the problem:** Solve the equation $$2|x^2 - 2| + 5 = 6|2x^2 - 3|.$$\n\n2. **Understand the absolute value expressions:** The equation involves absolute values of quadratic expressions. We need to consider cases based on the expressions inside the absolute values.\n\n3. **Define variables for simplicity:** Let $$a = x^2.$$ Then the equation becomes $$2|a - 2| + 5 = 6|2a - 3|.$$\n\n4. **Consider cases for $$|a - 2|$$:**\n- Case 1: $$a - 2 \geq 0 \Rightarrow a \geq 2,$$ so $$|a - 2| = a - 2.$$\n- Case 2: $$a - 2 < 0 \Rightarrow a < 2,$$ so $$|a - 2| = 2 - a.$$\n\n5. **Consider cases for $$|2a - 3|$$:**\n- Case A: $$2a - 3 \geq 0 \Rightarrow a \geq \frac{3}{2},$$ so $$|2a - 3| = 2a - 3.$$\n- Case B: $$2a - 3 < 0 \Rightarrow a < \frac{3}{2},$$ so $$|2a - 3| = 3 - 2a.$$\n\n6. **Analyze combined cases:** We have four combined cases based on $$a$$ intervals:\n- Case 1A: $$a \geq 2$$ and $$a \geq \frac{3}{2}$$ (so $$a \geq 2$$)\n- Case 1B: $$a \geq 2$$ and $$a < \frac{3}{2}$$ (impossible since $$a \geq 2 > \frac{3}{2}$$)\n- Case 2A: $$a < 2$$ and $$a \geq \frac{3}{2}$$ (so $$\frac{3}{2} \leq a < 2$$)\n- Case 2B: $$a < 2$$ and $$a < \frac{3}{2}$$ (so $$a < \frac{3}{2}$$)\n\n7. **Solve each valid case:**\n\n**Case 1A: $$a \geq 2$$**\nEquation: $$2(a - 2) + 5 = 6(2a - 3)$$\nSimplify left: $$2a - 4 + 5 = 2a + 1$$\nRight: $$12a - 18$$\nEquation: $$2a + 1 = 12a - 18$$\nBring terms together: $$1 + 18 = 12a - 2a \Rightarrow 19 = 10a$$\nSolve for $$a$$: $$a = \frac{19}{10} = 1.9$$\nCheck domain: $$a \geq 2$$ but $$1.9 < 2$$, so no solution here.\n\n**Case 2A: $$\frac{3}{2} \leq a < 2$$**\nEquation: $$2(2 - a) + 5 = 6(2a - 3)$$\nLeft: $$4 - 2a + 5 = 9 - 2a$$\nRight: $$12a - 18$$\nEquation: $$9 - 2a = 12a - 18$$\nBring terms: $$9 + 18 = 12a + 2a \Rightarrow 27 = 14a$$\nSolve for $$a$$: $$a = \frac{27}{14} \approx 1.9286$$\nCheck domain: $$1.5 \leq 1.9286 < 2$$ valid, so solution here.\n\n**Case 2B: $$a < \frac{3}{2}$$**\nEquation: $$2(2 - a) + 5 = 6(3 - 2a)$$\nLeft: $$4 - 2a + 5 = 9 - 2a$$\nRight: $$18 - 12a$$\nEquation: $$9 - 2a = 18 - 12a$$\nBring terms: $$9 - 18 = -12a + 2a \Rightarrow -9 = -10a$$\nSolve for $$a$$: $$a = \frac{9}{10} = 0.9$$\nCheck domain: $$0.9 < 1.5$$ valid, so solution here.\n\n8. **Recall $$a = x^2$$, solve for $$x$$:**\n- From Case 2A: $$x^2 = \frac{27}{14}$$\n$$x = \pm \sqrt{\frac{27}{14}} = \pm \frac{3\sqrt{14}}{14}.$$\n- From Case 2B: $$x^2 = \frac{9}{10}$$\n$$x = \pm \sqrt{\frac{9}{10}} = \pm \frac{3}{\sqrt{10}} = \pm \frac{3\sqrt{10}}{10}.$$\n\n9. **Final solutions:**\n$$x = \pm \frac{3\sqrt{14}}{14}, \pm \frac{3\sqrt{10}}{10}.$$