Please help

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.