Find the positive root of $\sqrt{3x^2+6}=9$

The given equation is $\sqrt{3x^2+6}=9$

Squaring the given equation on both the sides

$\begin{array}{r}{\left(\sqrt{3x^2+6}\right)}^2=9^2\\ \Rightarrow 3x^2+6=81\\ \Rightarrow 3x^2=75\\ \Rightarrow x^2=25\\ \Rightarrow x=\pm 5\\ \end{array}$

Hence, the positive root of the given equation is 5.