To solve for x in the equation 2log3(-x)=1+log3(x+6), we can first use the properties of logarithms to simplify the equation.

Use the log rule log_a(x) + log_a(y) = log_a(xy)
2log3(-x) = log3((-x)^2)
2log3(-x) = log3(x^2)
-2x = x^2+6
0 = x^2 + 2x + 6

Now, we have a quadratic equation 0 = x^2 + 2x + 6. We can solve this using the quadratic formula:

x = [-b ± sqrt(b^2 - 4ac)] / 2a
x = [-2 ± sqrt(2^2 - 4(1)(6))] / 2(1)
x = [-2 ± sqrt(4 - 24)] / 2
x = [-1 ± sqrt(-20)] / 2
x = (-1 ± √20i) / 2

Therefore, the solutions for x are:

x = (-1 + √20i) / 2
x = (-1 - √20i) / 2

So, the solutions for x are complex numbers.