(2-3)^2-(4^2-3^2)

The answer to "(2 - 3)^2 - (4^2 - 3^2)" would be -6.

To solve "(2 - 3)^2 - (4^2 - 3^2)", we should start by simplifying everything in the parantheses. That way all of the numbers in the parantheses will be simplified into just one simple value.

Let's start with "(2 - 3)". We can subtract 2 and 3 easily, like so:

2 - 3 = -1

Remember that since 3 is being subtracted from 2, we would not have 1, but we would instead have -1 because 3 is greater than 2.

Now we have a number that we can plug into our equation in place of "(2 - 3)":

-1^2 - (4^2 - 3^2)

Now let's simplify "(4^2 - 3^2)". Here we need to find the squares of 4 and 3 and then subtract them. We can multiply the numbers by themselves to obtain:

4 * 4 = 16

and:

3 * 3 = 9

So now we can replace "4^2" with 16 and "3^2" with 9:

(16 - 9)

And from here we can easily subtract 9 from 16:

16 - 9 = 7

Now we can plug 7 into our original equation just as we did with -1:

-1^2 - 7

Next, we need to find "-1^2" so that we can subtract 7 from that to obtain our answer:

-1 * -1 = 1

Note that -1 * -1 would not be -1, because multiplying a negative number by a negative number will always give a positive number.

Now we can plug in 1 for "-1^2" to obtain "1 - 7". The final step is to subtract 7 from 1, giving:

1 - 7 = -6

So our answer is -6. I hope I helped.

-pokemonfan58

Another excellent explanation pokemonfan58.