Solve in $\mathbb{Q}$ the equation $x^2-(\sqrt{2}+1)x+\sqrt{2}=0$. Solve in $\mathbb{Q}$ the equation $x^2-(\sqrt{2}+1)x+\sqrt{2}=0$
Somebody can help me? I dont remember how to do.         

Using Quadratic Formula 
If $ax^2+bx+c=0\;,$ Then $\displaystyle x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Here $a = 1$ and $b=-\left(\sqrt{2}+1\right)$ and $c=\sqrt{2}$
So $$\displaystyle x = \frac{(\sqrt{2}+1)\pm \sqrt{2+1+2\sqrt{2}-4\sqrt{2}}}{2}$$
So $$\displaystyle x = \frac{(\sqrt{2}+1)\pm (\sqrt{2}-1)}{2} = \sqrt{2}\;\;,1$$
