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

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

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(-[3+(2-3)^2])/(3-2^2+2)

\frac{- [ 3 + ( 2 - 3 ) ^{2} ]}{3 - 2 ^{2} + 2}

- 4

Solution steps

\frac{- [ 3 + ( 2 - 3 ) ^{2} ]}{3 - 2 ^{2} + 2}

Solve

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

Follow the PEMDAS order of operations

Calculate within parentheses

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

3 + (2 - 3)^{2}

Calculate within parentheses

(2 - 3) : - 1

2 - 3

2 - 3 = - 1

= - 1

= 3 + (- 1)^{2}

Calculate exponents

(- 1)^{2} : 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

(- 1)^{2} = 1^{2} = 1

= 1

= 3 + 1

Add and subtract (left to right)

3 + 1 : 4

3 + 1

3 + 1 = 4

= 4

= 4

= - 4

Solve

3 - 2^{2} + 2 : 1

Follow the PEMDAS order of operations

Calculate exponents

2^{2} : 4

2^{2}

2^{2} = 4

= 4

= 3 - 4 + 2

Add and subtract (left to right)

3 - 4 + 2 : 1

3 - 4 + 2

3 - 4 = - 1

= - 1 + 2

- 1 + 2 = 1

= 1

= 1

= \frac{- 4}{1}

Apply rule

\frac{a}{1} = a

= - 4

1 2^{2} + 1 7^{2}

3^{0} + 3^{1}

3 \cdot 3 \cdot 3 \cdot 3 \cdot 3^{2}

- 5 + 2 (6 + 2^{2})

1 2^{2} + 2^{2}

What is (-[3+(2-3)^2])/(3-2^2+2) ?
The solution to (-[3+(2-3)^2])/(3-2^2+2) is -4