1. **State the problem:** Rationalize the denominator and simplify the expression $$\frac{14}{3-\sqrt{2}}$$. 2. **Formula and rule:** To rationalize a denominator with a binomial involving a square root, multiply numerator and denominator by the conjugate of the denominator. The conjugate of $$3-\sqrt{2}$$ is $$3+\sqrt{2}$$. 3. **Multiply numerator and denominator by the conjugate:** $$\frac{14}{3-\sqrt{2}} \times \frac{3+\sqrt{2}}{3+\sqrt{2}} = \frac{14(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$$ 4. **Simplify the denominator using difference of squares:** $$ (3-\sqrt{2})(3+\sqrt{2}) = 3^2 - (\sqrt{2})^2 = 9 - 2 = 7 $$ 5. **Expand the numerator:** $$ 14(3+\sqrt{2}) = 14 \times 3 + 14 \times \sqrt{2} = 42 + 14\sqrt{2} $$ 6. **Write the fraction:** $$ \frac{42 + 14\sqrt{2}}{7} $$ 7. **Simplify by dividing numerator terms by 7:** $$ \frac{\cancel{42}^6 + \cancel{14}^2\sqrt{2}}{\cancel{7}} = 6 + 2\sqrt{2} $$ **Final answer:** $$6 + 2\sqrt{2}$$