The equation \(x^2+kx=5x+3\) must have only one solution for the line \(y\) to be a tangent line to the quadratic function \(f(x)\).
\(x^2+kx=5x+3\\
x^2+kx-5x-3=0\\
x^2+(k-5)x-3=0\\
\Delta=(k-5)^2-4\cdot1\cdot(-3)=(k-5)^2+12\\\\
(k-5)^2+12=0\\
(k-5)^2=-12\\
k\in \emptyset\)
So there isn't such value of k.
