positive root of the equation $x^2+x-3-\sqrt{3}=0$ $x^2+x-3-\sqrt{3}=0$
using the quadratic formula we get 
$$x=\frac{-1+\sqrt{13+4\sqrt{3}}}{2}$$ for the positive root
but the actual answer is simply   $x=\sqrt3$
I am unable to perform the simplification any help would we helpful

$13+4\sqrt{3}=(2\sqrt{3})^2+2(2\sqrt{3})(1)+1=(2\sqrt{3}+1)^2$
So it is just $\dfrac{-1+2\sqrt{3}+1}{2}$.

We may also start from the original equation.
\begin{align*}
x^2+x-3-\sqrt{3}&=0\\
x^2-3+x-\sqrt{3}&=0\\
(x-\sqrt{3})(x+\sqrt{3})+x-\sqrt{3}&=0\\
(x-\sqrt{3})(x+\sqrt{3}+1)&=0
\end{align*}
