Solve sqrt{[(21/2-1/6+02)*9]-11/4}=

$$\sqrt { [ ( 2 \frac { 1 } { 2 } - \frac { 1 } { 6 } + 02 ) \cdot 9 ] - \frac { 11 } { 4 } } =$$

Evaluate

$$\frac{\sqrt{73}}{2}\approx 4.272001873$$

Solution Steps

Multiply 2 and 2 to get 4.

$$\sqrt{\left(\frac{4+1}{2}-\frac{1}{6}+0\times 2\right)\times 9-\frac{11}{4}}$$

Add 4 and 1 to get 5.

$$\sqrt{\left(\frac{5}{2}-\frac{1}{6}+0\times 2\right)\times 9-\frac{11}{4}}$$

Least common multiple of 2 and 6 is 6. Convert \frac{5}{2} and \frac{1}{6} to fractions with denominator 6.

$$\sqrt{\left(\frac{15}{6}-\frac{1}{6}+0\times 2\right)\times 9-\frac{11}{4}}$$

Since \frac{15}{6} and \frac{1}{6} have the same denominator, subtract them by subtracting their numerators.

$$\sqrt{\left(\frac{15-1}{6}+0\times 2\right)\times 9-\frac{11}{4}}$$

Subtract 1 from 15 to get 14.

$$\sqrt{\left(\frac{14}{6}+0\times 2\right)\times 9-\frac{11}{4}}$$

Reduce the fraction \frac{14}{6} to lowest terms by extracting and canceling out 2.

$$\sqrt{\left(\frac{7}{3}+0\times 2\right)\times 9-\frac{11}{4}}$$

Multiply 0 and 2 to get 0.

$$\sqrt{\left(\frac{7}{3}+0\right)\times 9-\frac{11}{4}}$$

Add \frac{7}{3} and 0 to get \frac{7}{3}.

$$\sqrt{\frac{7}{3}\times 9-\frac{11}{4}}$$

Express \frac{7}{3}\times 9 as a single fraction.

$$\sqrt{\frac{7\times 9}{3}-\frac{11}{4}}$$

Multiply 7 and 9 to get 63.

$$\sqrt{\frac{63}{3}-\frac{11}{4}}$$

Divide 63 by 3 to get 21.

$$\sqrt{21-\frac{11}{4}}$$

Convert 21 to fraction \frac{84}{4}.

$$\sqrt{\frac{84}{4}-\frac{11}{4}}$$

Since \frac{84}{4} and \frac{11}{4} have the same denominator, subtract them by subtracting their numerators.

$$\sqrt{\frac{84-11}{4}}$$

Subtract 11 from 84 to get 73.

$$\sqrt{\frac{73}{4}}$$

Rewrite the square root of the division \sqrt{\frac{73}{4}} as the division of square roots \frac{\sqrt{73}}{\sqrt{4}}.

$$\frac{\sqrt{73}}{\sqrt{4}}$$

Calculate the square root of 4 and get 2.

$$\frac{\sqrt{73}}{2}$$