Two point boundary value problem $x''+x =0$ Side conditions x(0)=0 and x(1)=1
I know that I need to find the roots first but don't know how to continue.
Using $x=e^{\lambda t}$, the roots are found by $\lambda^2 + 1 = 0$, which gives us $i$ as the root. 

Your roots are 
$$ \lambda = i,\quad \lambda =-i. $$
Now just construct the solution as
$$ x(t) =c_1e^{it} +c_2 e^{-it}. $$
You can simplify the above if you want. I think you know how to advance.
Added: As I said you can simplify the above answer to the form

$$ x(t) = A\sin(t) + B\cos(t). $$

