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}&amp;=0\\
x^2-3+x-\sqrt{3}&amp;=0\\
(x-\sqrt{3})(x+\sqrt{3})+x-\sqrt{3}&amp;=0\\
(x-\sqrt{3})(x+\sqrt{3}+1)&amp;=0
\end{align*}
Rather than pulling $\sqrt{13+4\sqrt3}=1+2\sqrt3$ magically out of a hat, write $\sqrt{13+4\sqrt3}=a+b\sqrt3,$

as suggested by Parcly Taxel in a comment.  Squaring both sides, $13+4\sqrt3=(a^2+3b^2)+2ab\sqrt3.$    

$13=a^2+3b^2$ and $4=2ab$ means $b=2/a$ and $13=a^2+\dfrac{12}{a^2},$ i.e., $a^4-13a^2+12=0$, i.e., $(a^2-1)(a^2-12)=0$, i.e., $a=\pm1$ or $a=\pm2\sqrt3$ and respectively $b=\pm2$ or $\pm\dfrac 1{\sqrt3}.  $  Note that $\sqrt{13+4\sqrt3}&gt;0.$  Can you take it from here?