Solve {2+[70div2+4-(4-2)^2+8]*3+2^4}

$$\{ 2 + [ 70 \div 2 + 4 - ( 4 - 2 ) ^ { 2 } + 8 ] \cdot 3 + 2 ^ { 4 } \}$$

Evaluate

$$147$$

Solution Steps

Divide 70 by 2 to get 35.

$$2+\left(35+4-\left(4-2\right)^{2}+8\right)\times 3+2^{4}$$

Add 35 and 4 to get 39.

$$2+\left(39-\left(4-2\right)^{2}+8\right)\times 3+2^{4}$$

Subtract 2 from 4 to get 2.

$$2+\left(39-2^{2}+8\right)\times 3+2^{4}$$

Calculate 2 to the power of 2 and get 4.

$$2+\left(39-4+8\right)\times 3+2^{4}$$

Subtract 4 from 39 to get 35.

$$2+\left(35+8\right)\times 3+2^{4}$$

Add 35 and 8 to get 43.

$$2+43\times 3+2^{4}$$

Multiply 43 and 3 to get 129.

$$2+129+2^{4}$$

Add 2 and 129 to get 131.

$$131+2^{4}$$

Calculate 2 to the power of 4 and get 16.

$$131+16$$

Add 131 and 16 to get 147.

$$147$$

Factor

$$3\times 7^{2}$$