1. **State the problem:** Find all real solutions to the equation $$100x^2 + 20x + 1 = 16.$$\n\n2. **Rewrite the equation:** Move all terms to one side to set the equation to zero:\n$$100x^2 + 20x + 1 - 16 = 0$$\nwhich simplifies to\n$$100x^2 + 20x - 15 = 0.$$\n\n3. **Identify the quadratic formula:** For a quadratic equation $$ax^2 + bx + c = 0,$$ the solutions are given by\n$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$$\nHere, $$a = 100,$$ $$b = 20,$$ and $$c = -15.$$\n\n4. **Calculate the discriminant:**\n$$\Delta = b^2 - 4ac = 20^2 - 4 \times 100 \times (-15) = 400 + 6000 = 6400.$$\nSince $$\Delta > 0,$$ there are two real solutions.\n\n5. **Find the square root of the discriminant:**\n$$\sqrt{6400} = 80.$$\n\n6. **Calculate the two solutions:**\n$$x = \frac{-20 \pm 80}{2 \times 100} = \frac{-20 \pm 80}{200}.$$\n\n- For the plus sign:\n$$x_1 = \frac{-20 + 80}{200} = \frac{60}{200} = 0.3.$$\n- For the minus sign:\n$$x_2 = \frac{-20 - 80}{200} = \frac{-100}{200} = -0.5.$$\n\n**Final answer:** The real solutions are $$x = 0.3, -0.5.$$