Simplify |(-2)^2+6(3-15)|+(-9) square root of 81*16

$| {(- 2)}^{2} + 6 (3 - 15) | + (- 9) \sqrt{81 \cdot 16}$

Step 1

Simplify each term.

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

$| 4 + 6 (3 - 15) | + (- 9) \sqrt{81 \cdot 16}$

Subtract $15$ from $3$ .

$| 4 + 6 \cdot - 12 | + (- 9) \sqrt{81 \cdot 16}$

Multiply $6$ by $- 12$ .

$| 4 - 72 | + (- 9) \sqrt{81 \cdot 16}$

Subtract $72$ from $4$ .

$| - 68 | + (- 9) \sqrt{81 \cdot 16}$

The absolute value is the distance between a number and zero. The distance between $- 68$ and $0$ is $68$ .

$68 + (- 9) \sqrt{81 \cdot 16}$

Multiply $81$ by $16$ .

$68 - 9 \sqrt{1296}$

Rewrite $1296$ as $36^{2}$ .

$68 - 9 \sqrt{36^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$68 - 9 \cdot 36$

Multiply $- 9$ by $36$ .

$68 - 324$

Step 2

Subtract $324$ from $68$ .

$- 256$