First thing we do is assume that #x != 0# and #x !=5#, because if it was either of those values we'd be dividing by 0 and that's not allowed. Since x is not 0, we can factor it on the numerator and cut it out with the one on the denominator.

#(cancel(x)* (45x^2 - 9x) * 2)/(cancel(x)*6(x-5))#

We can simplify that 2 on the top with that 6 on the bottom since #6 = 2*3#

#(cancel(2)* (45x^2 - 9x))/(cancel(2)*3(x-5))#

#45 = 9*5# and #9 = 9*1# so we can put 9 in evidence on the numerator and simplify with that 3 on the denominator

#(3 * cancel(3) * (5x^2 - x))/(cancel(3) * (x-5))#

Last but not least, we can put an x in evidence on the numerator

#(3x*(5x - 1))/(x-5)#

We can't really do anything more significant here, we could multiply the numerator so it's one equation but this way it's easier to identify the roots (there's only one, #1/5#, by the way, since we said way back in the beginning that x couldn't be 0)