# How do you evaluate 2+6(9-3^2)-2?

Mar 25, 2017

The answer is $0$.

#### Explanation:

Use the order of operations $\left(\text{PEMDAS}\right)$.

$\textcolor{red}{\text{P}}$ = parentheses
$\textcolor{b l u e}{\text{E}}$ = exponents
$\textcolor{\mathmr{and} a n \ge}{\text{MD}}$ = multiplying and dividing
$\textcolor{v i o \le t}{\text{AS}}$ = adding and subtracting

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$2 + 6 \textcolor{red}{\left(9 - {3}^{2}\right)} - 2 \textcolor{w h i t e}{\text{XXX}}$ Evaluate the $\textcolor{red}{\text{P}}$arentheses

within parentheses, we start over with $\text{PEMDAS}$

$\left(9 - {3}^{2}\right) \textcolor{w h i t e}{\text{XXXXXXX}}$ No $\textcolor{red}{\text{P}}$arentheses
$\left(9 - \textcolor{b l u e}{{3}^{2}}\right) \textcolor{w h i t e}{\text{XXxxxxxxx}}$ Evaluate the $\textcolor{b l u e}{\text{E}}$xponents
$\left(9 - 9\right) \textcolor{w h i t e}{\text{xxXXxxxxxx}}$ No $\textcolor{\mathmr{and} a n \ge}{\text{M}}$ultiplication or $\textcolor{\mathmr{and} a n \ge}{\text{D}}$ivision
$\left(\textcolor{v i o \le t}{9 - 9}\right) \textcolor{w h i t e}{\text{xxXXxxxxxx}}$ Evaluate the $\textcolor{v i o \le t}{\text{A}}$ddition and $\textcolor{v i o \le t}{\text{S}}$ubtraction
$= \left(0\right)$

$2 + 6 \left(0\right) - 2 \textcolor{w h i t e}{\text{XXXXXX}}$ No $\textcolor{b l u e}{\text{E}}$xponents

$2 + \textcolor{\mathmr{and} a n \ge}{6 \left(0\right)} - 2 \textcolor{w h i t e}{\text{XXXXXX}}$ Evaluate the $\textcolor{\mathmr{and} a n \ge}{\text{M}}$ultiplication and $\textcolor{\mathmr{and} a n \ge}{\text{D}}$ivision

$\textcolor{v i o \le t}{2 + 0 - 2} \textcolor{w h i t e}{\text{XXXXXXX/}}$ Evaluate the $\textcolor{v i o \le t}{\text{A}}$ddition and $\textcolor{v i o \le t}{\text{S}}$ubtraction

$= 0$