1. **State the problem:** We want to show that $$\frac{\sqrt{12}}{\sqrt{3}+2}$$ can be written in the form $$a + \sqrt{b}$$ where $$a$$ and $$b$$ are integers. 2. **Recall the formula and rule:** To simplify expressions with radicals in the denominator, multiply numerator and denominator by the conjugate of the denominator. The conjugate of $$\sqrt{3}+2$$ is $$2 - \sqrt{3}$$. 3. **Multiply numerator and denominator by the conjugate:** $$\frac{\sqrt{12}}{\sqrt{3}+2} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{\sqrt{12}(2 - \sqrt{3})}{(\sqrt{3}+2)(2 - \sqrt{3})}$$ 4. **Simplify the denominator using difference of squares:** $$ (\sqrt{3}+2)(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 $$ 5. **Simplify the numerator:** $$ \sqrt{12}(2 - \sqrt{3}) = 2\sqrt{12} - \sqrt{12}\sqrt{3} $$ 6. **Simplify radicals:** $$ \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} $$ 7. **Substitute back:** $$ 2\sqrt{12} - \sqrt{12}\sqrt{3} = 2(2\sqrt{3}) - (2\sqrt{3})(\sqrt{3}) = 4\sqrt{3} - 2\sqrt{3} \times \sqrt{3} $$ 8. **Simplify further:** $$ 4\sqrt{3} - 2 \times 3 = 4\sqrt{3} - 6 $$ 9. **Write the entire expression:** $$ \frac{4\sqrt{3} - 6}{1} = -6 + 4\sqrt{3} $$ 10. **Final form:** $$ a = -6, \quad b = 3 $$ Thus, $$\frac{\sqrt{12}}{\sqrt{3}+2} = -6 + 4\sqrt{3}$$ where $$a = -6$$ and $$b = 3$$ are integers.