Answer
$\left\{x\ |\ x \leq 1 \right\}$ or $\left(-\infty, \ 1 \right]$
Work Step by Step
$x(9x-5)\leq(3x-1)^{2}$
$ 9x^{2}-5x\leq 9x^{2}-6x+1\qquad$ ... add $ +6x-9x^{2}$
$x \leq 1$
Solution set: $\left\{x\ |\ x \leq 1 \right\}$ or $\left(-\infty, \ 1 \right]$
