Solve left(2+sqrt{6}right)left(3sqrt{2}-2sqrt{3}right)

$$\left( 2+ \sqrt{ 6 } \right) \left( 3 \sqrt{ 2 } -2 \sqrt{ 3 } \right)$$

Evaluate

$$2\sqrt{3}\approx 3.464101615$$

Solution Steps

Apply the distributive property by multiplying each term of 2+\sqrt{6} by each term of 3\sqrt{2}-2\sqrt{3}.

$$6\sqrt{2}-4\sqrt{3}+3\sqrt{6}\sqrt{2}-2\sqrt{6}\sqrt{3}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$6\sqrt{2}-4\sqrt{3}+3\sqrt{2}\sqrt{3}\sqrt{2}-2\sqrt{6}\sqrt{3}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$6\sqrt{2}-4\sqrt{3}+3\times 2\sqrt{3}-2\sqrt{6}\sqrt{3}$$

Multiply 3 and 2 to get 6.

$$6\sqrt{2}-4\sqrt{3}+6\sqrt{3}-2\sqrt{6}\sqrt{3}$$

Combine -4\sqrt{3} and 6\sqrt{3} to get 2\sqrt{3}.

$$6\sqrt{2}+2\sqrt{3}-2\sqrt{6}\sqrt{3}$$

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$6\sqrt{2}+2\sqrt{3}-2\sqrt{3}\sqrt{2}\sqrt{3}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$6\sqrt{2}+2\sqrt{3}-2\times 3\sqrt{2}$$

Multiply -2 and 3 to get -6.

$$6\sqrt{2}+2\sqrt{3}-6\sqrt{2}$$

Combine 6\sqrt{2} and -6\sqrt{2} to get 0.

$$2\sqrt{3}$$