Chapter 9 - Roots and Radicals - 9.5 - Solving Radical Equations - Problem Set 9.5 - Page 424: 29

$n=2$

Squaring both sides of the equation and then using the properties of equality, we obtain: \begin{array}{l}\require{cancel}\left( \sqrt{3n-4} \right)^2=\left( \sqrt{n} \right)^2 \\\\ 3n-4=n \\\\ 3n-n=4 \\\\ 2n=4 \\\\ n=\dfrac{4}{2} \\\\ n=2 .\end{array} Upon checking, $ n=2 $ satisfies the original equation.