The roots of a quadratic equation are \((3 - \sqrt{3})\) and \((3 + \sqrt{3})\). Find its equation.

The roots of a quadratic equation of the form \(ax^{2} + bx + c = 0\) are given by the formula: $$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$ If the roots are \((3 - \sqrt{3})\) and \((3 + \sqrt{3})\), then we have: $$x = 3 - \sqrt{3} \quad \text{or} \quad x = 3 + \sqrt{3}$$ Therefore, we can write the equation as: $$(x - (3 - \sqrt{3}))(x - (3 + \sqrt{3})) = 0$$ Expanding the brackets, we get: $$x^{2} - (3 - \sqrt{3} + 3 + \sqrt{3})x + (3 - \sqrt{3})(3 + \sqrt{3}) = 0$$ Simplifying, we get: $$x^{2} - 6x + 6 = 0$$ Hence, the equation of the quadratic with roots \((3 - \sqrt{3})\) and \((3 + \sqrt{3})\) is \(x^{2} - 6x + 6 = 0\). Therefore, the answer is.