1. **State the problem:** Solve the equation $$2^{2x} - 3 \cdot 2^{x+2} + 32 = 0$$. 2. **Rewrite the equation using properties of exponents:** Recall that $$2^{2x} = (2^x)^2$$ and $$2^{x+2} = 2^x \cdot 2^2 = 4 \cdot 2^x$$. 3. **Substitute:** Let $$y = 2^x$$, so the equation becomes: $$y^2 - 3 \cdot 4 y + 32 = 0$$ which simplifies to $$y^2 - 12 y + 32 = 0$$. 4. **Solve the quadratic equation:** Use the quadratic formula: $$y = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot 32}}{2} = \frac{12 \pm \sqrt{144 - 128}}{2} = \frac{12 \pm \sqrt{16}}{2}$$. 5. **Calculate the roots:** $$y = \frac{12 \pm 4}{2}$$ So, $$y_1 = \frac{12 + 4}{2} = 8$$ $$y_2 = \frac{12 - 4}{2} = 4$$. 6. **Back-substitute for $$x$$:** Recall $$y = 2^x$$, so $$2^x = 8$$ and $$2^x = 4$$. 7. **Solve for $$x$$:** $$2^x = 8 = 2^3 \implies x = 3$$ $$2^x = 4 = 2^2 \implies x = 2$$. **Final answer:** $$x = 2$$ or $$x = 3$$.