Solve 6sqrt{12}+sqrt{48}-2sqrt{75}=

$$6 \sqrt{ 12 } + \sqrt{ 48 } -2 \sqrt{ 75 } =$$

Evaluate

$$6\sqrt{3}\approx 10.392304845$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$6\times 2\sqrt{3}+\sqrt{48}-2\sqrt{75}$$

Multiply 6 and 2 to get 12.

$$12\sqrt{3}+\sqrt{48}-2\sqrt{75}$$

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}.

$$12\sqrt{3}+4\sqrt{3}-2\sqrt{75}$$

Combine 12\sqrt{3} and 4\sqrt{3} to get 16\sqrt{3}.

$$16\sqrt{3}-2\sqrt{75}$$

Factor 75=5^{2}\times 3. Rewrite the square root of the product \sqrt{5^{2}\times 3} as the product of square roots \sqrt{5^{2}}\sqrt{3}. Take the square root of 5^{2}.

$$16\sqrt{3}-2\times 5\sqrt{3}$$

Multiply -2 and 5 to get -10.

$$16\sqrt{3}-10\sqrt{3}$$

Combine 16\sqrt{3} and -10\sqrt{3} to get 6\sqrt{3}.

$$6\sqrt{3}$$