What is the following quotient? \( \frac{2-\sqrt{8}}{4+\sqrt{12}} \)

To simplify the quotient \( \frac{2-\sqrt{8}}{4+\sqrt{12}} \), first simplify the square roots: \[ \sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2} \] \[ \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3} \] Substituting these back into the expression gives: \[ \frac{2 - 2\sqrt{2}}{4 + 2\sqrt{3}} \] Next, we can factor out the common factors in both the numerator and denominator: \[ = \frac{2(1 - \sqrt{2})}{2(2 + \sqrt{3})} \] Now, cancel the common factor of 2: \[ = \frac{1 - \sqrt{2}}{2 + \sqrt{3}} \] To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, \(2 - \sqrt{3}\): \[ = \frac{(1 - \sqrt{2})(2 - \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})} \] Calculating the denominator: \[ (2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1 \] Now, simplify the numerator: \[ (1 - \sqrt{2})(2 - \sqrt{3}) = 2 - \sqrt{3} - 2\sqrt{2} + \sqrt{6} \] So the final result is: \[ = 2 - \sqrt{3} - 2\sqrt{2} + \sqrt{6} \] Thus, the simplified expression for the quotient \( \frac{2-\sqrt{8}}{4+\sqrt{12}} \) is: \[ \boxed{2 - \sqrt{3} - 2\sqrt{2} + \sqrt{6}} \]