What is 14-[2\times (8-(29-3\times 2^3))-4] ?

The solution to 14-[2\times (8-(29-3\times 2^3))-4] is 12

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14-[2\times (8-(29-3\times 2^3))-4]

14 - [2 \cdot (8 - (29 - 3 \cdot 2^{3})) - 4]

12

Solution steps

14 - [2 (8 - (29 - 3 \cdot 2^{3})) - 4]

Follow the PEMDAS order of operations

Calculate within parentheses

[2 \cdot (8 - (29 - 3 \cdot 2^{3})) - 4] : 2

2 (8 - (29 - 3 \cdot 2^{3})) - 4

Calculate within parentheses

(8 - (29 - 3 \cdot 2^{3})) : 3

8 - (29 - 3 \cdot 2^{3})

Calculate within parentheses

(29 - 3 \cdot 2^{3}) : 5

29 - 3 \cdot 2^{3}

Calculate exponents

2^{3} : 8

2^{3}

2^{3} = 8

= 8

= 29 - 3 \cdot 8

Multiply and divide (left to right)

3 \cdot 8 : 24

3 \cdot 8

3 \cdot 8 = 24

= 24

= 29 - 24

Add and subtract (left to right)

29 - 24 : 5

29 - 24

29 - 24 = 5

= 5

= 5

= 8 - 5

Add and subtract (left to right)

8 - 5 : 3

8 - 5

8 - 5 = 3

= 3

= 3

= 2 \cdot 3 - 4

Multiply and divide (left to right)

2 \cdot 3 : 6

2 \cdot 3

2 \cdot 3 = 6

= 6

= 6 - 4

Add and subtract (left to right)

6 - 4 : 2

6 - 4

6 - 4 = 2

= 2

= 2

= 14 - 2

Add and subtract (left to right)

14 - 2 : 12

14 - 2

14 - 2 = 12

= 12

= 12

4 \times (2 \times (- 1))

(- 5) + (+ 3) + (+ 4) + (- 2)

\frac{4}{3} + (\frac{5}{6} + \frac{1}{2})

- 4 - (3)^{2} + 9

6^{2} \div 3^{2}

What is 14-[2\times (8-(29-3\times 2^3))-4] ?
The solution to 14-[2\times (8-(29-3\times 2^3))-4] is 12