Solve sqrt{6}left(3sqrt{2}+5sqrt{3}right)-7sqrt{48}

$$\sqrt{ 6 } \left( 3 \sqrt{ 2 } +5 \sqrt{ 3 } \right) -7 \sqrt{ 48 }$$

Evaluate

$$15\sqrt{2}-22\sqrt{3}\approx -16.891914331$$

Solution Steps

Use the distributive property to multiply \sqrt{6} by 3\sqrt{2}+5\sqrt{3}.

$$3\sqrt{6}\sqrt{2}+5\sqrt{6}\sqrt{3}-7\sqrt{48}$$

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}.

$$3\sqrt{2}\sqrt{3}\sqrt{2}+5\sqrt{6}\sqrt{3}-7\sqrt{48}$$

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

$$3\times 2\sqrt{3}+5\sqrt{6}\sqrt{3}-7\sqrt{48}$$

Multiply 3 and 2 to get 6.

$$6\sqrt{3}+5\sqrt{6}\sqrt{3}-7\sqrt{48}$$

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}.

$$6\sqrt{3}+5\sqrt{3}\sqrt{2}\sqrt{3}-7\sqrt{48}$$

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

$$6\sqrt{3}+5\times 3\sqrt{2}-7\sqrt{48}$$

Multiply 5 and 3 to get 15.

$$6\sqrt{3}+15\sqrt{2}-7\sqrt{48}$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$6\sqrt{3}+15\sqrt{2}-7\times 4\sqrt{3}$$

Multiply -7 and 4 to get -28.

$$6\sqrt{3}+15\sqrt{2}-28\sqrt{3}$$

Combine 6\sqrt{3} and -28\sqrt{3} to get -22\sqrt{3}.

$$-22\sqrt{3}+15\sqrt{2}$$