How to factor the following: 60x^2-130x+60=0

I get the following but I do not know how to get rid of the 10.

60x^2-130x+60=0
10(6x^2-13x+6)= 0
10(6x^2-4x)-(9x+6)=0
10 2x(3x-2) - 3(3x-2)= 0
10(3x-2)(2x-3)=0
10 = 0 or 3x-2 +2 = 0 + 2 or 2x-3+3 = 0 +3
10 = 0 or x =2/3 or x = 3/2

How to get rid of the 10?

You've got it exactly right - x = 2/3 or 3/2. How can 10=0? It doesn't - you can just divide both sides by 10, and you're left with the simpler equation.

Actually,you factored it correctly and the following working that you did is correct:

60x^2-130x+60=0
10(6x^2-13x+6)= 0
10(6x^2-4x)-(9x+6)=0
10 2x(3x-2) - 3(3x-2)= 0
10(3x-2)(2x-3)=0

Don't try to get rid of the 10,as it is just the amount that the equation is multiplied by.Rather,you will just ignore the 10,since you will only need to care about the part which is (3x-2)(2x-3)=0.From there,you get x =2/3 or x = 3/2,as you've stated.

Lets start here: 10(3x-2)(2x-3)=0
All you do is divide both sides by 10; the left side divided by 10 ends up being (3x-2)(2x-3) and the right side ends up still being 0. So after this short step all you have left is (3x-2)(2x-3)=0. It's as simple as that! Hope that helped!