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Yeah, thats correct.

At the x value of n, n^2+2 has to be equal to 2n+5 for the function to be continuous at x=n.

\(n^2+2=2n+5\)

\(n^2-2n-3=0\)

\((n-3)(n+1)=0\)

\(n=3, n=-1\)

The sum of all values of n is 3-1=2.

I think you forgot to type in what the numbers sum to.

Actually Juriemagic, you can just post your image in imgur or imbb or something like that. Imgur and ImgBB are websites where you can upload any image saved to your PC. Then copy the direct link and paste it in when you are posting an image.

Thanks Guest!!!

Thank you!!!

Thanks all of yyou!!!

Thanks for the help!!!

Actually, this is a written contest where you have to do 30 questions in 40 minutes by hand, no calculator.