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I get 69 when I calculate this, but my calculator says it's 29. I've gone over this several times and can't figure out why I'm wrong. I'm also not entirely sure if I've used the calculator properly.

What answer is correct, and any idea where I'm heading wrong here?

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  • $\begingroup$ Okay so $(-2-2)^2$ gives me $8$. $-(-2+3)(-2-3)$ gives me $5$. And $-4(-2^2+2)$ gives me $8^2-8$ which is $56$. Add these together and I get $69$. $\endgroup$ Oct 12 '15 at 21:37
  • $\begingroup$ You didn't use the parentheses properly in the very first expression, $(-2-2)^2$ is not $8$. $\endgroup$
    – hardmath
    Oct 12 '15 at 21:40
  • $\begingroup$ I thought that gave me (4+4), am I supposed to solve the inside of the parantheses before I apply the exponentiation? Did not know that.. $\endgroup$ Oct 12 '15 at 21:41
  • $\begingroup$ Yes, that's the purpose of parentheses. $\endgroup$
    – hardmath
    Oct 12 '15 at 21:42
  • $\begingroup$ Also, It seems that you did like $-4\times (-2^2)=8^2$. This is wrong. Since $-2^2=-(2^2)=-4$, we have $-4\times (-2^2)=-4\times (-4)=16$. $\endgroup$
    – mathlove
    Oct 12 '15 at 21:45
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You have used you calculator properly. It is

$$(-2-2)^2=(-4)^2=16,$$ $$(-2+3)(-2-3)=1\cdot (-5)=-5,$$ and $$-4(-2^2+2)=-4(-4+2)=-4(-2)=8.$$ Thus, you get

$$16-(-5)+8=16+5+8=29.$$

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  • $\begingroup$ $-2^2$ is the same as -2 ⋅ -2, right? So shouldn't that be 4 and not -4? $\endgroup$ Oct 12 '15 at 21:53
  • $\begingroup$ Nevermind, I guess multiplication>subtraction applies here as well.. $\endgroup$ Oct 12 '15 at 21:55
  • $\begingroup$ Yes, if we want to express $(-2)^2$ we need parentheses to force the unary minus to be done first. $\endgroup$
    – hardmath
    Oct 13 '15 at 0:00
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I believe your calculator is correct.

\begin{align*} &(-2-2)^2-(-2+3)(-2-3)-4(-2^2+2)\\ &=(-4)^2-(1)(-5)-4(-4+2)\\ &=(16)-(-5)+16-8\\ &=16+5+16-8\\ &=21+8\\ &=29 \end{align*} Note that $-2^2=-4$ whereas $(-2)^2=4$.

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