I'm sorry math, but I'm not seeing how the quadratic formula has anthing to do with solving for k. Now, if we were solving for x then the quadratic formula would be most appropriate,
K is not quadratic at all. In fact it is linear.
But for x...
$\displaystyle (x-1)(-x+k)=\frac{1}{x-1}*(x-1)$
$\displaystyle -x^2+kx+x-k=1$
$\displaystyle -x^2+kx+x-k-1=0$
$\displaystyle x^2-kx-x+k+1=0$
$\displaystyle x^2-(k-1)x+(k+1)=0$
The quadratic formula states that
$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Let a=1, b=-(k-1), and c=(k+1)
You can do the rest...