Solve (sqrt{3}+4)(sqrt{6}-3)

$$( \sqrt { 3 } + 4 ) ( \sqrt { 6 } - 3 )$$

Evaluate

$$3\sqrt{2}+4\sqrt{6}-3\sqrt{3}-12\approx -3.155552764$$

Solution Steps

Apply the distributive property by multiplying each term of \sqrt{3}+4 by each term of \sqrt{6}-3.

$$\sqrt{3}\sqrt{6}-3\sqrt{3}+4\sqrt{6}-12$$

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}.

$$\sqrt{3}\sqrt{3}\sqrt{2}-3\sqrt{3}+4\sqrt{6}-12$$

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

$$3\sqrt{2}-3\sqrt{3}+4\sqrt{6}-12$$