To solve this equation, we can first make a substitution by letting y = lg(x-1).

The equation then becomes:
3y^2 - 10y + 3 = 0

This is a quadratic equation that we can solve using the quadratic formula:
y = (-(-10) ± √((-10)^2 - 433)) / 2*3
y = (10 ± √(100 - 36)) / 6
y = (10 ± √64) / 6
y = (10 ± 8) / 6

Therefore, y = 3 or y = 1/3

Now, substituting back in y = lg(x-1):
lg(x-1) = 3 or lg(x-1) = 1/3

For lg(x-1) = 3:
x-1 = 10^3
x-1 = 1000
x = 1001

For lg(x-1) = 1/3:
x-1 = 10^(1/3)
x-1 = 10^(1/3)
x = 1 + 10^(1/3)

Therefore, the solutions to the equation 3lg^2(x-1)-10lg(x-1)+3=0 are x = 1001 and x = 1 + 10^(1/3).