Evaluate |3-6|-(12÷6+1)^2

$| 3 - 6 | - {(12 \div 6 + 1)}^{2}$

Step 1

Simplify each term.

Subtract $6$ from $3$ .

$| - 3 | - {(12 \div 6 + 1)}^{2}$

The absolute value is the distance between a number and zero. The distance between $- 3$ and $0$ is $3$ .

$3 - {(12 \div 6 + 1)}^{2}$

Divide $12$ by $6$ .

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

Add $2$ and $1$ .

$3 - 3^{2}$

Raise $3$ to the power of $2$ .

$3 - 1 \cdot 9$

Multiply $- 1$ by $9$ .

$3 - 9$

Step 2

Subtract $9$ from $3$ .

$- 6$