<s>[INST] Find the quadratic equation equation of $x_1, x_2$. 
Let $x_1 = 1 + \sqrt 3, x_2 = 1-\sqrt 3$. Find the quadratic equation $ax^2+bx+c = 0$ which $x_1, x_2$ are it's solutions.  

By Vieta's theorem:  
$$x_1\cdot x_2 = \frac{c}{a} \implies c=-2a$$
$$x_1 + x_2 = -\frac{b}{a} \implies b = -2a$$
Therefore, $b=c$
So we have a quadratic equation with the form:
$$-\frac{b}{2}x^2 + bx + b = 0$$
Applying $x_1$ for our equation I get $b=0$. Why? 
The final answer is: $x^2-2x-2$.
 [/INST] From your equation:
$−\frac{b}{2}x^2+bx+b=0$
notice that it's equal to $b*(−\frac{1}{2}x^2+x+1)=0$
Multiplying both sides of the equation by -2 we get:
$b*(x^2-2x-2)=0$
So the solutions to this become either $b=0$
or $x^2-2x-2=0$
Which imply $x=1 \pm \sqrt3$
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