What is (3+2)^2+[3(4-6)^2] ?

The solution to (3+2)^2+[3(4-6)^2] is 37

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(3+2)^2+[3(4-6)^2]

(3 + 2)^{2} + [3 (4 - 6)^{2}]

37

Solution steps

(3 + 2)^{2} + [3 (4 - 6)^{2}]

Follow the PEMDAS order of operations

Calculate within parentheses

(3 + 2) : 5

3 + 2

3 + 2 = 5

= 5

= 5^{2} + [3 (4 - 6)^{2}]

Calculate within parentheses

[3 (4 - 6)^{2}] : 12

3 (4 - 6)^{2}

Calculate within parentheses

(4 - 6) : - 2

4 - 6

4 - 6 = - 2

= - 2

= 3 (- 2)^{2}

Calculate exponents

(- 2)^{2} : 4

(- 2)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2} = 4

= 4

= 3 \cdot 4

Multiply and divide (left to right)

3 \cdot 4 : 12

3 \cdot 4

3 \cdot 4 = 12

= 12

= 12

= 5^{2} + 12

Calculate exponents

5^{2} : 25

5^{2}

5^{2} = 25

= 25

= 25 + 12

Add and subtract (left to right)

25 + 12 : 37

25 + 12

25 + 12 = 37

= 37

= 37

(44 + 20) \div 4 - 2 (9 \div 3) - 2 [18 + 3 (13 - 9) - 5]

16 \div 2 + 8 \times 2 - 3 \times (5 - 2)

- 1000000 + 760299 + 27769

3 (2)^{2} - 5 (2)

(76 - 4^{2}) \div (14 - 2)