Line touches a curve. Find a value of $x$ in the quadratic curve The line $y = 2x - 3$ touches the curve $y = x^2 + kx + 6$
Find the possible values of $k$.
I would like a tip towards solving the problem rather than the entire answer. I tried comparing them to each other but ended up with $x^2 - 2x + 9 = kx$ and did not know what to do next.

We can also take an "inside=out" approach.  We want the line $ \ y \ = \ 2x - 3 \ \ $ to be tangent to the parabola, so the parabola will need to be in contact at the one point on the curve where its first derivative ("slope") is equal to $ \ 2 \ \ . $  The problem is that varying the middle coefficient in $ \ y \ = \ x^2 + kx + 6 \ \ $ is not a simple translation of the parabola in the plane, so we need to work out what happens to that supposed tangent point.
We will need to have $ \ y' \ = \ 2x + k \ = \ 2 \ \Rightarrow \ x \ = \ \frac{2 - k}{2} \ \ . $  For the parabola to possess such a tangent point on the line $ \ y + 3 \ = \ 2x \ \ , $ we then have
$$ \left[ \ \left(\frac{2 - k}{2} \right)^2 \ + \ k·\left(\frac{2 - k}{2} \right) \ + \ 6 \ \right] \ + \ \ 3 \ \ = \ \ 2·\left(\frac{2 - k}{2} \right)   $$
$$ \Rightarrow \ \ ( 2 - k   )^2 \ + \ 2k· ( 2 - k ) \ + \ 36 \ \  = \ \ 4· ( 2 - k ) \ \ \Rightarrow \ \ k^2 \ - \ 4k \ - \ 32 \ \ = \ \ 0 \ \  . $$
This provides the two values of $ \ k \ $ without needing to be concerned about the number of intersections between the line and the parabola.  Should we be interested in the locations of the tangent points, we have the relation between $ \ k \ $ and the $ \ x-$coordinate, and either curve equation will provide the $ \ y-$coordinate. The graph below shows the two solution parabolas.

