Please help, I tried

The left hand side of the expression is 1 if (i) $x^2 -5x+2=0$   or  (ii) $x^2 - 4x + 2 = 1$

These are quadratics that can be solved using the standard quadratic formula.  They will give values of x in the form $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ 

Adding these two values will cancel the discriminant part leaving you with a sum of $\frac{-b}{a}$.  

Do this for each quadratic and then add the results.

Also if $x^2 - 4x + 2 = -1$ and $x^2 -5x+2$ is even then the RHS will be 1  (This will give you two more x values).