Chapter 9 - Roots and Radicals - 9.5 - Solving Radical Equations - Problem Set 9.5 - Page 424: 30

$n=\dfrac{1}{3}$

Squaring both sides of the equation and then using the properties of equality, we obtain: \begin{array}{l}\require{cancel}\left( \sqrt{5n-1} \right)^2=\left( \sqrt{2n} \right)^2 \\\\ 5n-1=2n \\\\ 5n-2n=1 \\\\ 3n=1 \\\\ n=\dfrac{1}{3} .\end{array} Upon checking, $ n=\dfrac{1}{3} $ satisfies the original equation.