How do you evaluate #-3+6(1-3)^2#?

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Answer 1 · 2018-05-31T12:41:33

See a solution process below:

First, execute the Subtraction operation within the Parenthesis:

#-3 + 6(color(red)(1) - color(red)(3))^2 => -3 + 6(-2)^2#

Next, execute the Exponent operation:

#-3 + 6(-2)^color(red)(2) => -3 + 6 xx 4#

Then, execute the Multiplication operation:

#-3 + color(red)(6) xx color(red)(4) => -3 + 24#

Now, execute the Additionoperation:

#color(red)(-3) + color(red)(24) => 21#

Answer 2 · 2018-06-06T09:13:44

#21#

Count the number of terms first and simplify each term to a single value. These are added or subtracted in the last line.

#color(blue)(-3)" + "color(green)(6(1-3)^2)#

Within each term do brackets first.
Then powers and roots
Then multiply and divide,

#=color(blue)(-3)" + "color(green)(6(-2)^2)" "larr# brackets

#=color(blue)(-3)" + "color(green)(6(4)" "larr# powers

#=color(blue)(-3)" + "color(green)(24)" "larr# muliplication

#= 21" "larr# addition