Solve 40-2lfloor3+8+(12+3)+4-(8-6)rfloor-4(1-8)=

$$40-2 \lfloor 3+8+(12+3)+4-(8-6) \rfloor -4(1-8)=$$

Evaluate

$$12$$

Solution Steps

Add 3 and 8 to get 11.

$$40-2\lfloor 11+12+3+4-\left(8-6\right)\rfloor -4\left(1-8\right)$$

Add 11 and 12 to get 23.

$$40-2\lfloor 23+3+4-\left(8-6\right)\rfloor -4\left(1-8\right)$$

Add 23 and 3 to get 26.

$$40-2\lfloor 26+4-\left(8-6\right)\rfloor -4\left(1-8\right)$$

Add 26 and 4 to get 30.

$$40-2\lfloor 30-\left(8-6\right)\rfloor -4\left(1-8\right)$$

Subtract 6 from 8 to get 2.

$$40-2\lfloor 30-2\rfloor -4\left(1-8\right)$$

Subtract 2 from 30 to get 28.

$$40-2\lfloor 28\rfloor -4\left(1-8\right)$$

The floor of a real number a is the largest integer number less than or equal to a. The floor of 28 is 28.

$$40-2\times 28-4\left(1-8\right)$$

Multiply 2 and 28 to get 56.

$$40-56-4\left(1-8\right)$$

Subtract 56 from 40 to get -16.

$$-16-4\left(1-8\right)$$

Subtract 8 from 1 to get -7.

$$-16-4\left(-7\right)$$

Multiply 4 and -7 to get -28.

$$-16-\left(-28\right)$$

The opposite of -28 is 28.

$$-16+28$$

Add -16 and 28 to get 12.

$$12$$

Factor

$$2^{2}\times 3$$