Solve sqrt{64}-3[(2-5)^2/3-4(3/2-10/4)^2]+3(7/9)

$$\sqrt { 64 } - 3 [ \frac { ( 2 - 5 ) ^ { 2 } } { 3 } - 4 ( \frac { 3 } { 2 } - \frac { 10 } { 4 } ) ^ { 2 } ] + 3 ( \frac { 7 } { 9 } )$$

Evaluate

$$\frac{40}{3}\approx 13.333333333$$

Solution Steps

Calculate the square root of 64 and get 8.

$$8-3\left(\frac{\left(2-5\right)^{2}}{3}-4\left(\frac{3}{2}-\frac{10}{4}\right)^{2}\right)+3\times \frac{7}{9}$$

Subtract 5 from 2 to get -3.

$$8-3\left(\frac{\left(-3\right)^{2}}{3}-4\left(\frac{3}{2}-\frac{10}{4}\right)^{2}\right)+3\times \frac{7}{9}$$

Calculate -3 to the power of 2 and get 9.

$$8-3\left(\frac{9}{3}-4\left(\frac{3}{2}-\frac{10}{4}\right)^{2}\right)+3\times \frac{7}{9}$$

Divide 9 by 3 to get 3.

$$8-3\left(3-4\left(\frac{3}{2}-\frac{10}{4}\right)^{2}\right)+3\times \frac{7}{9}$$

Reduce the fraction \frac{10}{4} to lowest terms by extracting and canceling out 2.

$$8-3\left(3-4\left(\frac{3}{2}-\frac{5}{2}\right)^{2}\right)+3\times \frac{7}{9}$$

Subtract \frac{5}{2} from \frac{3}{2} to get -1.

$$8-3\left(3-4\left(-1\right)^{2}\right)+3\times \frac{7}{9}$$

Calculate -1 to the power of 2 and get 1.

$$8-3\left(3-4\times 1\right)+3\times \frac{7}{9}$$

Multiply 4 and 1 to get 4.

$$8-3\left(3-4\right)+3\times \frac{7}{9}$$

Subtract 4 from 3 to get -1.

$$8-3\left(-1\right)+3\times \frac{7}{9}$$

Multiply 3 and -1 to get -3.

$$8-\left(-3\right)+3\times \frac{7}{9}$$

The opposite of -3 is 3.

$$8+3+3\times \frac{7}{9}$$

Add 8 and 3 to get 11.

$$11+3\times \frac{7}{9}$$

Multiply 3 and \frac{7}{9} to get \frac{7}{3}.

$$11+\frac{7}{3}$$

Add 11 and \frac{7}{3} to get \frac{40}{3}.

$$\frac{40}{3}$$

Factor

$$\frac{2 ^ {3} \cdot 5}{3} = 13\frac{1}{3} = 13.333333333333334$$