Solve 3^2+2^3[sqrt{36}(2^2-3)(2^2+1)-20]

$$3 ^ { 2 } + 2 ^ { 3 } [ \sqrt { 36 } ( 2 ^ { 2 } - 3 ) ( 2 ^ { 2 } + 1 ) - 20 ]$$

Evaluate

$$89$$

Solution Steps

Calculate 3 to the power of 2 and get 9.

$$9+2^{3}\left(\sqrt{36}\left(2^{2}-3\right)\left(2^{2}+1\right)-20\right)$$

Calculate 2 to the power of 3 and get 8.

$$9+8\left(\sqrt{36}\left(2^{2}-3\right)\left(2^{2}+1\right)-20\right)$$

Calculate the square root of 36 and get 6.

$$9+8\left(6\left(2^{2}-3\right)\left(2^{2}+1\right)-20\right)$$

Calculate 2 to the power of 2 and get 4.

$$9+8\left(6\left(4-3\right)\left(2^{2}+1\right)-20\right)$$

Subtract 3 from 4 to get 1.

$$9+8\left(6\times 1\left(2^{2}+1\right)-20\right)$$

Multiply 6 and 1 to get 6.

$$9+8\left(6\left(2^{2}+1\right)-20\right)$$

Calculate 2 to the power of 2 and get 4.

$$9+8\left(6\left(4+1\right)-20\right)$$

Add 4 and 1 to get 5.

$$9+8\left(6\times 5-20\right)$$

Multiply 6 and 5 to get 30.

$$9+8\left(30-20\right)$$

Subtract 20 from 30 to get 10.

$$9+8\times 10$$

Multiply 8 and 10 to get 80.

$$9+80$$

Add 9 and 80 to get 89.

$$89$$

Factor

$$89$$