Answer
$$1 < x < 5$$
Work Step by Step
$$\eqalign{
& {x^2} - 6x + 5 < 0 \cr
& {\text{Factoring}} \cr
& \left( {x - 5} \right)\left( {x - 1} \right) < 0 \cr
& {\text{The roots of the corresponding equation }}\left( {x - 5} \right)\left( {x - 1} \right){\text{ are}} \cr
& x = 1{\text{ and }}x = 5.{\text{ These roots partition the number line into}} \cr
& {\text{three intervals:}}\,\,\,\left( { - \infty ,1} \right),\,\,\left( {1,5} \right){\text{ and }}\left( {5,\infty } \right).{\text{ }} \cr
& {\text{Testing the intervals:}} \cr
& {\text{*For the interval }}\left( { - \infty ,1} \right){\text{ use the test value }}x = 0 \cr
& \left( {0 - 5} \right)\left( {0 - 1} \right) = 5,\,\,\,positive \cr
& {\text{*For the interval }}\left( {1,5} \right){\text{ use the test value }}x = 2 \cr
& \left( {2 - 5} \right)\left( {2 - 1} \right) = - 3,\,\,\,negative \cr
& {\text{*For the interval }}\left( {5,\infty } \right){\text{ use the test value }}x = 6 \cr
& \left( {6 - 5} \right)\left( {6 - 1} \right) = 5,\,\,positive \cr
& \cr
& {\text{Therefore, }}\left( {x - 5} \right)\left( {x - 1} \right) < 0{\text{ on the interval }}\left( {1,5} \right) \cr
& {\text{The solution set of the inequailty }}{x^2} - 6x + 5 < 0{\text{ is}} \cr
& \left( {1,5} \right),\,\,\,or\,\,\,\,1 < x < 5 \cr
& \cr
& {\text{Graph}} \cr} $$
