What is 2^2+(2\times 2)-8 ?

The solution to 2^2+(2\times 2)-8 is 0

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2^2+(2\times 2)-8

2^{2} + (2 \cdot 2) - 8

0

Solution steps

2^{2} + (2 \cdot 2) - 8

Follow the PEMDAS order of operations

Calculate within parentheses

(2 \cdot 2) : 4

2 \cdot 2

2 \cdot 2 = 4

= 4

= 2^{2} + 4 - 8

Calculate exponents

2^{2} : 4

2^{2}

2^{2} = 4

= 4

= 4 + 4 - 8

Add and subtract (left to right)

4 + 4 - 8 : 0

4 + 4 - 8

4 + 4 = 8

= 8 - 8

8 - 8 = 0

= 0

= 0

\frac{( \frac{1 + 6}{+ 3} )}{( 6 - 5 )}

\frac{\times ( - 4 )}{\times 3 - 16}

1 + 4 \times 4 \div 2

2 \times (11 - 3) + 7^{2}

(2)^{2} + 4 (2)