1. **State the problem:** Solve the quadratic equation $$x^2 - (2\sqrt{3} + 1)x + 2\sqrt{3} = 0$$. 2. **Recall the quadratic formula:** For an equation $$ax^2 + bx + c = 0$$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. 3. **Identify coefficients:** Here, $$a = 1$$, $$b = -(2\sqrt{3} + 1)$$, and $$c = 2\sqrt{3}$$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-(2\sqrt{3} + 1))^2 - 4 \times 1 \times 2\sqrt{3}$$ $$= (2\sqrt{3} + 1)^2 - 8\sqrt{3}$$ 5. **Expand and simplify:** $$(2\sqrt{3} + 1)^2 = (2\sqrt{3})^2 + 2 \times 2\sqrt{3} \times 1 + 1^2 = 4 \times 3 + 4\sqrt{3} + 1 = 12 + 4\sqrt{3} + 1 = 13 + 4\sqrt{3}$$ 6. **Substitute back:** $$\Delta = 13 + 4\sqrt{3} - 8\sqrt{3} = 13 - 4\sqrt{3}$$ 7. **Calculate the roots:** $$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{2\sqrt{3} + 1 \pm \sqrt{13 - 4\sqrt{3}}}{2}$$ 8. **Final answer:** $$x = \frac{2\sqrt{3} + 1 \pm \sqrt{13 - 4\sqrt{3}}}{2}$$ This gives the two solutions to the quadratic equation.