What is 5(4-2)^2+1^2(2^3-5)^2 ?

The solution to 5(4-2)^2+1^2(2^3-5)^2 is 29

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5(4-2)^2+1^2(2^3-5)^2

5 (4 - 2)^{2} + 1^{2} (2^{3} - 5)^{2}

29

Solution steps

5 (4 - 2)^{2} + 1^{2} (2^{3} - 5)^{2}

Follow the PEMDAS order of operations

Calculate within parentheses

(4 - 2) : 2

4 - 2

4 - 2 = 2

= 2

= 5 \cdot 2^{2} + 1^{2} (2^{3} - 5)^{2}

Calculate within parentheses

(2^{3} - 5) : 3

2^{3} - 5

Calculate exponents

2^{3} : 8

2^{3}

2^{3} = 8

= 8

= 8 - 5

Add and subtract (left to right)

8 - 5 : 3

8 - 5

8 - 5 = 3

= 3

= 3

= 5 \cdot 2^{2} + 1^{2} \cdot 3^{2}

Calculate exponents

2^{2} : 4

2^{2}

2^{2} = 4

= 4

= 5 \cdot 4 + 1^{2} \cdot 3^{2}

Calculate exponents

1^{2} : 1

1^{2}

1^{2} = 1

= 1

= 5 \cdot 4 + 1 \cdot 3^{2}

Calculate exponents

3^{2} : 9

3^{2}

3^{2} = 9

= 9

= 5 \cdot 4 + 1 \cdot 9

Multiply and divide (left to right)

5 \cdot 4 : 20

5 \cdot 4

5 \cdot 4 = 20

= 20

= 20 + 1 \cdot 9

Multiply and divide (left to right)

1 \cdot 9 : 9

1 \cdot 9

1 \cdot 9 = 9

= 9

= 20 + 9

Add and subtract (left to right)

20 + 9 : 29

20 + 9

20 + 9 = 29

= 29

= 29

\frac{120}{4}

7 \cdot (6 + 2)^{2} - 3^{2}

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

\frac{216}{3}

\frac{128}{4}