Take out the common factor 3: 5x3-34x2+69x-36.
Start by finding rational zeroes. To do this 36 would be the product of the three zeroes, that is, the cubic can be expressed in the form (5x-a)(x-b)(x-c) producing -abc as the constant term. A cubic has at least one real zero.
36=2232, so it has only two prime factors: 2 and 3. It's possible that one or both of these are factors in either positive or negative form.
Try x=2 first: cubic evaluates to 6, reject this. Try x=3 next: cubic evaluates to 0, so x=3 is a zero. Divide by this zero using synthetic division:
3 | 5 -34 69 -36
5 15 -47 | 36
5 -19 12 | 0 = 5x2-19x+12=(5x-4)(x-3),
which gives us two more real zeroes: x=⅘ and 3.
So we have a duplicate zero (x=3) and x=⅘.