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: $x^2-2x-2$.

Here's how I would do the "calculate and check" steps:
$$\begin{align}(x-x_1)(x-x_2)&=0\\
x_1&=1+\sqrt 3\\
x_2&=1-\sqrt 3\\
x^2-(x_1+x_2)x+x_1x_2&=0\\
x_1+x_2&=2\\
x_1x_2&=1^2-\sqrt 3^2=-2\\
x^2-2x-2&=0\end{align}$$
Now, taking the form that you created, we have
$$-\frac b2x^2+bx+b = 0$$
Plugging in $x=x_1=1+\sqrt 3$ results in
$$\begin{align}-\frac b2(1^2+2\sqrt 3+3)+b(1+\sqrt 3)+b&=0\tag 1\\
-\frac b2(4+2\sqrt 3)+2b+b\sqrt 3&=0\\
-2b-b\sqrt 3+2b+b\sqrt3 &= 0\\
0 &= 0\end{align}$$
Note that using $x=x_2=1-\sqrt 3$ simply places a couple negative signs so we end up with $(1^2-2\sqrt 3+3)$ and $(1-\sqrt 3)$ in $(1)$.
This check step should always result in $0=0$ when the equation and other calculations are done correctly.
