Solve sqrt{12/3}-2sqrt{45}+2sqrt{20}

$$\sqrt { 1 \frac { 2 } { 3 } } - 2 \sqrt { 45 } + 2 \sqrt { 20 }$$

Evaluate

$$\frac{\sqrt{15}}{3}-2\sqrt{5}\approx -3.181141506$$

Short Solution Steps

Multiply 1 and 3 to get 3.

$$\sqrt{\frac{3+2}{3}}-2\sqrt{45}+2\sqrt{20}$$

Add 3 and 2 to get 5.

$$\sqrt{\frac{5}{3}}-2\sqrt{45}+2\sqrt{20}$$

Rewrite the square root of the division \sqrt{\frac{5}{3}} as the division of square roots \frac{\sqrt{5}}{\sqrt{3}}.

$$\frac{\sqrt{5}}{\sqrt{3}}-2\sqrt{45}+2\sqrt{20}$$

Rationalize the denominator of \frac{\sqrt{5}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{\sqrt{5}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-2\sqrt{45}+2\sqrt{20}$$

The square of \sqrt{3} is 3.

$$\frac{\sqrt{5}\sqrt{3}}{3}-2\sqrt{45}+2\sqrt{20}$$

To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.

$$\frac{\sqrt{15}}{3}-2\sqrt{45}+2\sqrt{20}$$

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}.

$$\frac{\sqrt{15}}{3}-2\times 3\sqrt{5}+2\sqrt{20}$$

Multiply -2 and 3 to get -6.

$$\frac{\sqrt{15}}{3}-6\sqrt{5}+2\sqrt{20}$$

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}.

$$\frac{\sqrt{15}}{3}-6\sqrt{5}+2\times 2\sqrt{5}$$

Multiply 2 and 2 to get 4.

$$\frac{\sqrt{15}}{3}-6\sqrt{5}+4\sqrt{5}$$

Combine -6\sqrt{5} and 4\sqrt{5} to get -2\sqrt{5}.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply -2\sqrt{5} times \frac{3}{3}.

$$\frac{\sqrt{15}}{3}+\frac{3\left(-2\right)\sqrt{5}}{3}$$

Since \frac{\sqrt{15}}{3} and \frac{3\left(-2\right)\sqrt{5}}{3} have the same denominator, add them by adding their numerators.

$$\frac{\sqrt{15}+3\left(-2\right)\sqrt{5}}{3}$$

Do the multiplications in \sqrt{15}+3\left(-2\right)\sqrt{5}.

$$\frac{\sqrt{15}-6\sqrt{5}}{3}$$