I know that the quadratic factorises as $(x-(1+\sqrt{2}))$ times
something. So perhaps I could write the quadratic as
$$(x-(1+\sqrt{2}))(ax+b)$$
and work out what $a$ and $b$ could be.
Alternatively, perhaps I could think about the relation between
$x=\sqrt{2}$ and $x^2-2=0$, and see where that leads me to if I start
with $x=1+\sqrt{2}$ instead.
As a third possible approach, I could think about the graph of
$y=x^2-2$, which has $x=\sqrt{2}$ as a solution, and wonder what I
would have to do to get $x=1+\sqrt{2}$ as a solution instead.