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

The solution to (1)^2+4(1)-5 is 0

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

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

0

Solution steps

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

Follow the PEMDAS order of operations

Calculate exponents

(1)^{2} : 1

(1)^{2}

(1)^{2} = 1

= 1

= 1 + 4 (1) - 5

Multiply and divide (left to right)

4 (1) : 4

4 (1)

Apply rule:

(a) = a

(1) = 1

= 4 \cdot 1

Multiply the numbers:

4 \cdot 1 = 4

= 4

= 1 + 4 - 5

Add and subtract (left to right)

1 + 4 - 5 : 0

1 + 4 - 5

1 + 4 = 5

= 5 - 5

5 - 5 = 0

= 0

= 0

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

13 - 3 [8 + (4 - 9)]

4 + 5 \times 2 - 3 \times 3 + 12 \div 3

56 + [70 + (45 - 12) - 56 + 2]

2^{200} + 2^{200} + 2^{200} + 2^{200}