Answer
$\left\{\dfrac{2\pm\sqrt{5}}{5}\right\}$
Work Step by Step
In the form $x^2+bx=c,$ the given equation, $
25n^2-20n=1
,$ is equivalent to \begin{align*}\require{cancel} \dfrac{25n^2-20n}{25}&=\dfrac{1}{25} \\\\ \dfrac{\cancel{25}n^2}{\cancel{25}}-\dfrac{\cancelto4{20}n}{\cancelto5{25}}&=\dfrac{1}{25} \\\\ n^2-\dfrac{4n}{5}&=\dfrac{1}{25} .\end{align*} To complete the square of the expression at the left side, add $\left(\dfrac{b}{2}\right)^2$ to both sides of the equation. That is, \begin{align*} n^2-\dfrac{4n}{5}+\left(\dfrac{-4/5}{2}\right)^2&=\dfrac{1}{25}+\left(\dfrac{-4/5}{2}\right)^2 \\\\ n^2-\dfrac{4n}{5}+\left(\dfrac{-2}{5}\right)^2&=\dfrac{1}{25}+\left(\dfrac{-2}{5}\right)^2 \\\\ n^2-\dfrac{4n}{5}+\dfrac{4}{25}&=\dfrac{1}{25}+\dfrac{4}{25} \\\\ n^2-\dfrac{4n}{5}+\dfrac{4}{25}&=\dfrac{5}{25} .\end{align*} Using $a^2\pm2ab+b^2=(a\pm b)^2$ or the factoring of perfect square trinomials, the equation above is equivalent to \begin{align*} \left(n-\dfrac{2}{5}\right)^2&=\dfrac{5}{25} .\end{align*} Taking the square root of both sides (Square Root Principle), the equation above is equivalent to \begin{align*} n-\dfrac{2}{5}&=\pm\sqrt{\dfrac{5}{25}} \\\\ n-\dfrac{2}{5}&=\pm\sqrt{\dfrac{1}{25}\cdot5} \\\\ n-\dfrac{2}{5}&=\pm\sqrt{\left(\dfrac{1}{5}\right)^2\cdot5} \\\\ n-\dfrac{2}{5}&=\pm\dfrac{1}{5}\sqrt{5} \\\\ n-\dfrac{2}{5}&=\pm\dfrac{\sqrt{5}}{5} .\end{align*} Using the properties of equality, the equation above is equivalent to \begin{align*} n&=\dfrac{2}{5}\pm\dfrac{\sqrt{5}}{5} \\\\ n&=\dfrac{2\pm\sqrt{5}}{5} .\end{align*} Hence, the solution set of the equation $ 25n^2-20n=1 $ is $\left\{\dfrac{2\pm\sqrt{5}}{5}\right\}$.
