Solve (2^5+sqrt{36})*sqrt{2^2+72:6}=

$$( 2 ^ { 5 } + \sqrt { 36 } ) \cdot \sqrt { 2 ^ { 2 } + 72 : 6 } =$$

Evaluate

$$152$$

Solution Steps

Calculate 2 to the power of 5 and get 32.

$$\left(32+\sqrt{36}\right)\sqrt{2^{2}+\frac{72}{6}}$$

Calculate the square root of 36 and get 6.

$$\left(32+6\right)\sqrt{2^{2}+\frac{72}{6}}$$

Add 32 and 6 to get 38.

$$38\sqrt{2^{2}+\frac{72}{6}}$$

Calculate 2 to the power of 2 and get 4.

$$38\sqrt{4+\frac{72}{6}}$$

Divide 72 by 6 to get 12.

$$38\sqrt{4+12}$$

Add 4 and 12 to get 16.

$$38\sqrt{16}$$

Calculate the square root of 16 and get 4.

$$38\times 4$$

Multiply 38 and 4 to get 152.

$$152$$

Factor

$$2^{3}\times 19$$