$y=-2x+k$ and curve $y=\frac{2}{x-3}$ A straight line has equation $y=-2x+k$, where $k$ is a constant, and a curve has an equation $y=\frac{2}{x-3}$.
i)Show that the $x$-coordinates of any points of intersection of the line and curve are given by the equation $2x^2-(6+k)x+(2+3k)=0$
ii)Find the two values of $k$ for which the line is a tangent to the curve. 
The two tangents, given by the values of $k$ found in part ii), touch the curve at points.
iii)Find the coordinates of $A$ and $B$ and the equation of the line $AB$.
My attempt,
$-2x+k=\frac{2}{x-3}$
$(-2x+k)(x-3)=2$
Lastly, I got $2x^2-(6+k)x+3k+2=0$(Shown)
$b^2-4ac=0$
$(6+k)^2-4(2)(3k+2)=0$
$36+12k+k^2-24k-16=0$
$k^2-12k+20=0$
$k=2$ and $k=10$
How to proceed?

In this answer, I suppose that $A,B$ are the points where the line $y=-2x+k$ is tangent to the curve.
For $k=2$, $2x^2-(6+k)x+3k+2=0\Rightarrow 2x^2-8x+8=0,$ i.e. $2(x-2)^2=0$. So, $x=2$, and $y=\frac{2}{x-3}=\frac{2}{2-3}=-2$. So, we know that $(2,-2)$ is the point where the line $y=-2x+2$ is tangent to the curve. 
For $k=10$, $2x^2-(6+k)x+3k+2=0\Rightarrow 2x^2-16x+32=0,$ i.e. $2(x-4)^2=0$. So, $x=4$, and $y=\frac{2}{x-3}=\frac{2}{4-3}=2$. So, we know that $(4,2)$ is the point where the line $y=-2x+10$ is tangent to the curve.
Hence, the equation of the line $AB$ is 
$$y-(-2)=\frac{2-(-2)}{4-2}(x-2)\Rightarrow y=2x-6.$$
