General

Lol is this algebra 1 ?

As with all solution set problems, you have to find all the values of x that satisfy the equation.

You cannot combine the separate functions bound by the absolute value signs.
You can rewrite the absolute value functions into a set of positive and negative functions that when taken as a step function are equivalent to the original absolute value.

Breaking it down:
/x+2/ can be rewritten as
{(x+2)
{-(x+2)
which can be simplified as
{x+2
{-x-2
The values of x that satisfy this step function, which are x = 2, -2, also satisfy the starting absolute value function /x+2/

Going back to your original problem, you can rewrite it similarly to the below set of functions, and from there you can solve it using previously learned algebra methods.
(x+2) = 10 - (2x^2 -9)
-(x+2) = 10 - (2x^2 -9)
(x+2) = 10 + (2x^2 -9)
-(x+2) = 10 + (2x^2 -9)

Edit: nvm, made another mistake

ill fix it in a bit

Honestly, this is one of the more useless maths I never had to use again.

I can't remember what you do but I think it was taking turns.
So let's call one of the absolute A and the other B.

A=+
B=+

A=+
B=-

A=-
B=+

A=-
B=-

Here's a link with the same type of question. I didn't read it so don't trust me.
http://www.purplemath.com/modules/solveabs2.htm