<s>[INST] Re-write a quadratic equation in another form? $x^2 + \sqrt{2}x = \frac{1}{2}$
I need to find the real solutions for this equation and write it in this form:  
$$\frac{-\sqrt{A} \pm B}{C}$$
So when I work the problem out with the quadratic equation I get: $x = 0.118121$
I had no idea how to even put that in the form described so I assumed they didn't want me to solve but just put it in that form so I put it as follows:
$A = x^2$
$B = \sqrt{2}x$
$C = \frac{1}{2}$
That was wrong also. I'm not really sure what they're asking me to do. I know the answer is:  $x = 0.118121$ but is it even possible to put it in the form described? Any help would be appreciated. 
 [/INST] Form of any quadratic Equation is $ax^2+bx+c=0$
Here for you $ a=1 , b= \sqrt{2}$ and $c = -\frac{1}{2}$,  $$x^2 + \sqrt{2}x + \left(-\frac{1}{2} \right)=0$$
Solution is $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-\sqrt{2}\pm \sqrt{(\sqrt{2})^2-4\cdot 1 \cdot \left(-\frac{1}{2}\right)}}{2 \cdot 1}$$
$$x=\frac{-\sqrt{2}\pm \sqrt{2+2}}{2 }$$
$$x=\frac{-\sqrt{2}\pm \sqrt{4}}{2 }$$
$$x=\frac{-\sqrt{2}\pm 2}{2 }$$
So $$A=2 ,B =2 ,C=2$$
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