Show that the line $x+y=q$ intersects the ellipse $x^2-2x+2y^2=3$ at two different points if $q^2<2q+5$ What should be the proper way to answer this question?
My working is as follows, but is it the right way to answer the question, since it seems like I am only showing that $q^2<2q+5$?
$x+y=q \tag{1}$
$x^2-2x+2y^2=3 \tag{2}$
From $(1)$, $y=q-x\tag{3}$
Substitute $(3)$ into $(2)$,
$\begin{align}x^2-2x+2(q-x)^2&=3\\x^2-2x+2(q^2-2qx+x^2)&=3\\3x^2-(2+4q)x+2q^2-3&=0\end{align}$
Since the line intersects the ellipse at two different points, therefore $b^2-4ac>0$
$(-2-4q)^2-4(3)(2q^2-3)>0$
$4+16q+16q^2-24q^2+36>0$
$8q^2-16q-40<0$
$q^2-2q-5<0$
$q^2<2q+5$
Does my working answer the quesiton, or are there more accurate approaches to solve this question?

Your calculations are correct. Here you can find also for what values of $q$ the line $x+y=q$ and the ellpise $x^2-2x+2y^2=3$ have two points in common.
First, you can rewrite the equation of the ellipse in the form:
$$(x-1)^2+2y^2=4$$
Now, substituing $y=q-x$, you have:
$$(x-1)^2+2(q-x)^2=4\leftrightarrow 3x^2-2x(2q+1)+2q^2-3=0\leftrightarrow x=\frac{2(2q+1)\pm\sqrt{4(2q+1)^2-12(2q^2-3)}}{6}$$
The $|Delta$ must be greater then $0$ and so:
$$q^2-2q-5<0 \leftrightarrow q<1-\sqrt{6} \lor q>1+\sqrt{6}$$
