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?