jmc

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, $k=\boxed{-5}$.