Solve 2^{2}-[2-(3-2^2*sqrt{9}]+[sqrt{16}-(24div2^2)]^5-2^2=

$$2 ^ { 2 } - [ 2 - ( 3 - 2 ^ { 2 } \cdot \sqrt { 9 } ] + [ \sqrt { 16 } - ( 24 \div 2 ^ { 2 } ) ] ^ { 5 } - 2 ^ { 2 } =$$

Evaluate

$$29$$

Solution Steps

Calculate 2 to the power of 2 and get 4.

$$4-\left(2-\left(3-2^{2}\sqrt{9}\right)+\left(\sqrt{16}-\frac{24}{2^{2}}\right)^{5}-2^{2}\right)$$

Calculate 2 to the power of 2 and get 4.

$$4-\left(2-\left(3-4\sqrt{9}\right)+\left(\sqrt{16}-\frac{24}{2^{2}}\right)^{5}-2^{2}\right)$$

Calculate the square root of 9 and get 3.

$$4-\left(2-\left(3-4\times 3\right)+\left(\sqrt{16}-\frac{24}{2^{2}}\right)^{5}-2^{2}\right)$$

Multiply 4 and 3 to get 12.

$$4-\left(2-\left(3-12\right)+\left(\sqrt{16}-\frac{24}{2^{2}}\right)^{5}-2^{2}\right)$$

Subtract 12 from 3 to get -9.

$$4-\left(2-\left(-9\right)+\left(\sqrt{16}-\frac{24}{2^{2}}\right)^{5}-2^{2}\right)$$

The opposite of -9 is 9.

$$4-\left(2+9+\left(\sqrt{16}-\frac{24}{2^{2}}\right)^{5}-2^{2}\right)$$

Add 2 and 9 to get 11.

$$4-\left(11+\left(\sqrt{16}-\frac{24}{2^{2}}\right)^{5}-2^{2}\right)$$

Calculate the square root of 16 and get 4.

$$4-\left(11+\left(4-\frac{24}{2^{2}}\right)^{5}-2^{2}\right)$$

Calculate 2 to the power of 2 and get 4.

$$4-\left(11+\left(4-\frac{24}{4}\right)^{5}-2^{2}\right)$$

Divide 24 by 4 to get 6.

$$4-\left(11+\left(4-6\right)^{5}-2^{2}\right)$$

Subtract 6 from 4 to get -2.

$$4-\left(11+\left(-2\right)^{5}-2^{2}\right)$$

Calculate -2 to the power of 5 and get -32.

$$4-\left(11-32-2^{2}\right)$$

Subtract 32 from 11 to get -21.

$$4-\left(-21-2^{2}\right)$$

Calculate 2 to the power of 2 and get 4.

$$4-\left(-21-4\right)$$

Subtract 4 from -21 to get -25.

$$4-\left(-25\right)$$

The opposite of -25 is 25.

$$4+25$$

Add 4 and 25 to get 29.

$$29$$

Factor

$$29$$