1. **State the problem:** Solve for $x$ in the equation $$2^{2x+1} - 5 \times 2^x - 3 = 0.$$\n\n2. **Rewrite the equation:** Let $y = 2^x$. Then the equation becomes $$2^{2x+1} - 5 \times 2^x - 3 = 0 \implies 2 \times (2^x)^2 - 5 \times 2^x - 3 = 0,$$ which is $$2y^2 - 5y - 3 = 0.$$\n\n3. **Solve the quadratic equation:** Use the quadratic formula $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=-5$, and $c=-3$.\n\nCalculate the discriminant: $$\Delta = (-5)^2 - 4 \times 2 \times (-3) = 25 + 24 = 49.$$\n\nSo, $$y = \frac{5 \pm \sqrt{49}}{2 \times 2} = \frac{5 \pm 7}{4}.$$\n\n4. **Find the roots:**\n- For the plus sign: $$y = \frac{5 + 7}{4} = \frac{12}{4} = 3.$$\n- For the minus sign: $$y = \frac{5 - 7}{4} = \frac{-2}{4} = -\frac{1}{2}.$$\n\nSince $y = 2^x$ and $2^x > 0$ for all real $x$, discard $y = -\frac{1}{2}$.\n\n5. **Solve for $x$:**\n$$2^x = 3 \implies x = \log_2 3 = \frac{\log 3}{\log 2}.$$\n\n6. **Final answer:**\n$$x = \frac{\log 3}{\log 2} \approx 1.585.$$