If D is positive we have the result as I've given above (with 109 not 10-9). If D is negative, then srqt(D) = sqrt(|D|)*i, where i is sqrt(-1). The terms inside the absolute value then become, say k*i, where k is a real number. The absolute value of k*i is given by sqrt(k2), namely just k. so |ln(...)sqrt(D)| =|ln(...)|*sqrt(|D|). D can't be negative though, because if it were we would have D*|ln(...)|*sqrt(|D|) = - |D|*|ln(...)|*sqrt(|D|) = a negative number; but the LHS is a positive number, so D must be positive.
Here's a graph calculated by Mathcad (which is quite happy with imaginary numbers) for a range of values of D:

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