First, we can combine the logarithms on the left side using the product rule for logarithms:

lgx + lg(x+1) = lg(x(x+1)) = lg(x^2 + x)

Now, we have:

lg(x^2 + x) = lg(2x^2 - 6)

Since the bases of the logarithms are the same, we can drop the logarithms and set the expressions inside them equal to each other:

x^2 + x = 2x^2 - 6

Rearranging the terms gives us a quadratic equation:

x^2 - x - 6 = 0

Now we can factor this quadratic equation:

(x-3)(x+2) = 0

Setting each factor equal to zero gives us the possible values of x:

x - 3 = 0 or x + 2 = 0

x = 3 or x = -2

Therefore, the solutions to the equation lgx + lg(x+1) = lg(2x^2-6) are x = 3 and x = -2.