How do you evaluate #(4+ 6) \times 3+ 7- 12\times 8=#?

The evaluation of this expression begins with the rules for order of operations:
P - Parenthesis ( )
E - Exponents
M - Multiplication
D - Division
A - Addition
S - Subtraction
Moving left to right

#(4+6)*3+7-12*8#

Begin by evaluatiing the parenthesis

#10*3+7-12*8#

Now do the multiplication

#30+7-96#

Now add and subtract

#37-96#

#-59#

P arentheses
E xponents
M ultiply
D ivide
A dd
S ubtract
Exponents: The number of times you multiply the number by itself, in #4^2#, the exponent is #2#.

First you eliminate the parentheses by simplifying the expression between/in them.

#4+6= 10#

Now, after eliminating the parentheses, the expression will be:

#10⋅3+7−12⋅8=?#

There are no exponents in this equation so we re going to skip that.

You will be multiplying twice in this equation. #10⋅3# and #12⋅8#.
You are going to replace each of them with their answer.
You are pretty much eliminating the multiplication signs by simplifying them by finding the products.

Now, after finding the products multiplication signs, the expression will be:

#30+7-96#.

There is not going to be dividing in this expression. (there is no division sign)

Now you move on to the adding part.
There is only one addition sign.
Therefore, you add now!
#30+7=37#.

You are pretty much eliminating the addition sign by simplifying it
.
Now, after eliminating the addition sign, the expression will be:
#37-96#.

Lastly, you subtract! Since #96# is bigger than #37#, your answer will be negative.

Like if you have a loan of #96#, and you pay back #37#, you still have #59# more to pay back, so it is #-59#!
(I just told you a way to think about it.)

#37-96=-59#

Always remember the method, PEMDAS.

Source

I hope that helps!