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).
