Solve {3[(2-5)^2-4(3/2-10/4)^3]+3(7/9)}-sqrt{49}

$$\{ 3 [ ( 2 - 5 ) ^ { 2 } - 4 ( \frac { 3 } { 2 } - \frac { 10 } { 4 } ) ^ { 3 } ] + 3 ( \frac { 7 } { 9 } ) \} - \sqrt { 49 }$$

Evaluate

$$\frac{103}{3}\approx 34.333333333$$

Solution Steps

Subtract 5 from 2 to get -3.

$$3\left(\left(-3\right)^{2}-4\left(\frac{3}{2}-\frac{10}{4}\right)^{3}\right)+3\times \frac{7}{9}-\sqrt{49}$$

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

$$3\left(9-4\left(\frac{3}{2}-\frac{10}{4}\right)^{3}\right)+3\times \frac{7}{9}-\sqrt{49}$$

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

$$3\left(9-4\left(\frac{3}{2}-\frac{5}{2}\right)^{3}\right)+3\times \frac{7}{9}-\sqrt{49}$$

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

$$3\left(9-4\left(-1\right)^{3}\right)+3\times \frac{7}{9}-\sqrt{49}$$

Calculate -1 to the power of 3 and get -1.

$$3\left(9-4\left(-1\right)\right)+3\times \frac{7}{9}-\sqrt{49}$$

Multiply 4 and -1 to get -4.

$$3\left(9-\left(-4\right)\right)+3\times \frac{7}{9}-\sqrt{49}$$

The opposite of -4 is 4.

$$3\left(9+4\right)+3\times \frac{7}{9}-\sqrt{49}$$

Add 9 and 4 to get 13.

$$3\times 13+3\times \frac{7}{9}-\sqrt{49}$$

Multiply 3 and 13 to get 39.

$$39+3\times \frac{7}{9}-\sqrt{49}$$

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

$$39+\frac{7}{3}-\sqrt{49}$$

Add 39 and \frac{7}{3} to get \frac{124}{3}.

$$\frac{124}{3}-\sqrt{49}$$

Calculate the square root of 49 and get 7.

$$\frac{124}{3}-7$$

Subtract 7 from \frac{124}{3} to get \frac{103}{3}.

$$\frac{103}{3}$$

Factor

$$\frac{103}{3} = 34\frac{1}{3} = 34.333333333333336$$