Simplify (( square root of 225-11)*12)/(-12-(8-6^2))-|-3|

$\frac{(\sqrt{225} - 11) \cdot 12}{- 12 - (8 - 6^{2})} - | - 3 |$

Step 1

Simplify each term.

Simplify the numerator.

Rewrite $225$ as $15^{2}$ .

$\frac{(\sqrt{15^{2}} - 11) \cdot 12}{- 12 - (8 - 6^{2})} - | - 3 |$

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

$\frac{(15 - 11) \cdot 12}{- 12 - (8 - 6^{2})} - | - 3 |$

Subtract $11$ from $15$ .

$\frac{4 \cdot 12}{- 12 - (8 - 6^{2})} - | - 3 |$

Simplify the denominator.

Simplify each term.

Raise $6$ to the power of $2$ .

$\frac{4 \cdot 12}{- 12 - (8 - 1 \cdot 36)} - | - 3 |$

Multiply $- 1$ by $36$ .

$\frac{4 \cdot 12}{- 12 - (8 - 36)} - | - 3 |$

Subtract $36$ from $8$ .

$\frac{4 \cdot 12}{- 12 - - 28} - | - 3 |$

Multiply $- 1$ by $- 28$ .

$\frac{4 \cdot 12}{- 12 + 28} - | - 3 |$

Add $- 12$ and $28$ .

$\frac{4 \cdot 12}{16} - | - 3 |$

Multiply $4$ by $12$ .

$\frac{48}{16} - | - 3 |$

Divide $48$ by $16$ .

$3 - | - 3 |$

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

$3 - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$3 - 3$

Step 2

Subtract $3$ from $3$ .

$0$