Solve sqrt{36}-2*(4/3)^2+(2/3)

$$\sqrt { 36 } - 2 \times ( \frac { 4 } { 3 } ) ^ { 2 } + ( \frac { 2 } { 3 } )$$

Evaluate

$$\frac{28}{9}\approx 3.111111111$$

Solution Steps

Calculate the square root of 36 and get 6.

$$6-2\times \left(\frac{4}{3}\right)^{2}+\frac{2}{3}$$

Calculate \frac{4}{3} to the power of 2 and get \frac{16}{9}.

$$6-2\times \frac{16}{9}+\frac{2}{3}$$

Express 2\times \frac{16}{9} as a single fraction.

$$6-\frac{2\times 16}{9}+\frac{2}{3}$$

Multiply 2 and 16 to get 32.

$$6-\frac{32}{9}+\frac{2}{3}$$

Convert 6 to fraction \frac{54}{9}.

$$\frac{54}{9}-\frac{32}{9}+\frac{2}{3}$$

Since \frac{54}{9} and \frac{32}{9} have the same denominator, subtract them by subtracting their numerators.

$$\frac{54-32}{9}+\frac{2}{3}$$

Subtract 32 from 54 to get 22.

$$\frac{22}{9}+\frac{2}{3}$$

Least common multiple of 9 and 3 is 9. Convert \frac{22}{9} and \frac{2}{3} to fractions with denominator 9.

$$\frac{22}{9}+\frac{6}{9}$$

Since \frac{22}{9} and \frac{6}{9} have the same denominator, add them by adding their numerators.

$$\frac{22+6}{9}$$

Add 22 and 6 to get 28.

$$\frac{28}{9}$$

Factor

$$\frac{2 ^ {2} \cdot 7}{3 ^ {2}} = 3\frac{1}{9} = 3.111111111111111$$