Question:
For what negative value of $k$ is there exactly one solution to the system of equations \begin{align*}
y &= 2x^2 + kx + 6 \\
y &= -x + 4?
\end{align*}

Answer:
Setting the two expressions for $y$ equal to each other, it follows that $2x^2 + kx + 6 = -x + 4$. Re-arranging, $2x^2 + (k+1)x + 2 = 0$. For there to be exactly one solution for $x$, then the discriminant of the given quadratic must be equal to zero. Thus, $(k+1)^2 - 4 \cdot 2 \cdot 2 = (k+1)^2 - 16 = 0$, so $k+1 = \pm 4$. Taking the negative value, $k = \boxed{-5}$.