I have looked over this question several times, and I only understand the solution up to a point.
Solve the equation for $x$:
$$\ln x+\ln(x-1)=1 $$
First thing I do is apply the additive rule of logs
$$\ln(x(x-1))=1$$ $$\ln(x^2-x)=1$$ $$e^1=x^2-x$$ then setting up for quadratic $$ 0=x^2-x-e$$
Now here is where I get lost: defining the values of $a, b$, and $c$ for the quadratic. $$\begin{align*}a&=x^2\\ b&=-x\\ c&=-e \end{align*}$$
But the solution shows:
$$a=1, b=-1, c=-e$$
I am not sure the reasoning behind using these values for the quadratic?