State whether the quadratic equation $(x-\sqrt{2})^{2}-2(x+1)=0$ has two distinct real roots. Justify your answer.

(A) Given equation is $(x-\sqrt{2})^{2}-2(x+1)=0$.
Expanding the equation: $x^{2}-2\sqrt{2}x+2-2x-2=0$.
Simplifying: $x^{2}-(2+2\sqrt{2})x=0$.
Comparing with $ax^{2}+bx+c=0$,we get $a=1, b=-(2+2\sqrt{2}), c=0$.
The discriminant $D = b^{2}-4ac = (-(2+2\sqrt{2}))^{2}-4(1)(0)$.
$D = (2+2\sqrt{2})^{2} = 4 + 8 + 8\sqrt{2} = 12+8\sqrt{2}$.
Since $D > 0$,the quadratic equation has two distinct real roots.