Solve (2sqrt{20}-sqrt{45}+3sqrt{18})+(sqrt{72}-sqrt{80})

$$( 2 \sqrt { 20 } - \sqrt { 45 } + 3 \sqrt { 18 } ) + ( \sqrt { 72 } - \sqrt { 80 } )$$

Evaluate

$$3\sqrt{5}\left(\sqrt{10}-1\right)\approx 14.504999503$$

Solution Steps

Factor 20=2^{2}\times 5. Rewrite the square root of the product \sqrt{2^{2}\times 5} as the product of square roots \sqrt{2^{2}}\sqrt{5}. Take the square root of 2^{2}.

$$2\times 2\sqrt{5}-\sqrt{45}+3\sqrt{18}+\sqrt{72}-\sqrt{80}$$

Multiply 2 and 2 to get 4.

$$4\sqrt{5}-\sqrt{45}+3\sqrt{18}+\sqrt{72}-\sqrt{80}$$

Factor 45=3^{2}\times 5. Rewrite the square root of the product \sqrt{3^{2}\times 5} as the product of square roots \sqrt{3^{2}}\sqrt{5}. Take the square root of 3^{2}.

$$4\sqrt{5}-3\sqrt{5}+3\sqrt{18}+\sqrt{72}-\sqrt{80}$$

Combine 4\sqrt{5} and -3\sqrt{5} to get \sqrt{5}.

$$\sqrt{5}+3\sqrt{18}+\sqrt{72}-\sqrt{80}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$\sqrt{5}+3\times 3\sqrt{2}+\sqrt{72}-\sqrt{80}$$

Multiply 3 and 3 to get 9.

$$\sqrt{5}+9\sqrt{2}+\sqrt{72}-\sqrt{80}$$

Factor 72=6^{2}\times 2. Rewrite the square root of the product \sqrt{6^{2}\times 2} as the product of square roots \sqrt{6^{2}}\sqrt{2}. Take the square root of 6^{2}.

$$\sqrt{5}+9\sqrt{2}+6\sqrt{2}-\sqrt{80}$$

Combine 9\sqrt{2} and 6\sqrt{2} to get 15\sqrt{2}.

$$\sqrt{5}+15\sqrt{2}-\sqrt{80}$$

Factor 80=4^{2}\times 5. Rewrite the square root of the product \sqrt{4^{2}\times 5} as the product of square roots \sqrt{4^{2}}\sqrt{5}. Take the square root of 4^{2}.

$$\sqrt{5}+15\sqrt{2}-4\sqrt{5}$$

Combine \sqrt{5} and -4\sqrt{5} to get -3\sqrt{5}.

$$-3\sqrt{5}+15\sqrt{2}$$