Solve (2/3)^2-2*(-4/3)=

$$( \frac { 2 } { 3 } ) ^ { 2 } - 2 \cdot ( \frac { - 4 } { 3 } ) =$$

Evaluate

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

Solution Steps

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

$$\frac{4}{9}-2\times \frac{-4}{3}$$

Fraction \frac{-4}{3} can be rewritten as -\frac{4}{3} by extracting the negative sign.

$$\frac{4}{9}-2\left(-\frac{4}{3}\right)$$

Express 2\left(-\frac{4}{3}\right) as a single fraction.

$$\frac{4}{9}-\frac{2\left(-4\right)}{3}$$

Multiply 2 and -4 to get -8.

$$\frac{4}{9}-\frac{-8}{3}$$

Fraction \frac{-8}{3} can be rewritten as -\frac{8}{3} by extracting the negative sign.

$$\frac{4}{9}-\left(-\frac{8}{3}\right)$$

The opposite of -\frac{8}{3} is \frac{8}{3}.

$$\frac{4}{9}+\frac{8}{3}$$

Least common multiple of 9 and 3 is 9. Convert \frac{4}{9} and \frac{8}{3} to fractions with denominator 9.

$$\frac{4}{9}+\frac{24}{9}$$

Since \frac{4}{9} and \frac{24}{9} have the same denominator, add them by adding their numerators.

$$\frac{4+24}{9}$$

Add 4 and 24 to get 28.

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

Factor

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