1. **State the problem:** Find the value of $x$ that satisfies the equation $$5^{2x} - 6(5^x) + 5 = 0.$$\n\n2. **Rewrite the equation:** Notice that $5^{2x} = (5^x)^2$. Let $y = 5^x$. Then the equation becomes $$y^2 - 6y + 5 = 0.$$\n\n3. **Solve the quadratic equation:** Use the quadratic formula $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-6$, and $c=5$.\nCalculate the discriminant: $$\Delta = (-6)^2 - 4 \times 1 \times 5 = 36 - 20 = 16.$$\n\n4. **Find the roots:** $$y = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2}.$$\nSo, $$y_1 = \frac{6 + 4}{2} = 5,$$ $$y_2 = \frac{6 - 4}{2} = 1.$$\n\n5. **Back-substitute for $x$:** Recall $y = 5^x$. So,\nFor $y_1 = 5$: $$5^x = 5 \implies x = 1.$$\nFor $y_2 = 1$: $$5^x = 1 \implies x = 0$$ (since $5^0 = 1$).\n\n6. **Final answer:** The values of $x$ that satisfy the equation are $$\boxed{0 \text{ and } 1}.$$