1. **State the problem:** Solve the equation $$(2^x)^2 + 2^x - 6 = 0$$ for $x$. 2. **Rewrite the equation:** Note that $$(2^x)^2 = 2^{2x}$$, so the equation becomes: $$2^{2x} + 2^x - 6 = 0$$ 3. **Substitution:** Let $$y = 2^x$$. Then the equation becomes: $$y^2 + y - 6 = 0$$ 4. **Solve the quadratic equation:** Use the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=1$$, and $$c=-6$$. Calculate the discriminant: $$\Delta = 1^2 - 4 \times 1 \times (-6) = 1 + 24 = 25$$ So, $$y = \frac{-1 \pm \sqrt{25}}{2} = \frac{-1 \pm 5}{2}$$ 5. **Find the roots:** - $$y_1 = \frac{-1 + 5}{2} = \frac{4}{2} = 2$$ - $$y_2 = \frac{-1 - 5}{2} = \frac{-6}{2} = -3$$ 6. **Back-substitute for $x$:** Recall $$y = 2^x$$. - For $$y_1 = 2$$: $$2^x = 2$$ Since $$2^1 = 2$$, we have $$x = 1$$. - For $$y_2 = -3$$: $$2^x = -3$$ This is impossible because $$2^x > 0$$ for all real $$x$$. 7. **Final solution:** $$\boxed{x = 1}$$