Easier way to find amount of solutions between a line and quadratic?

A simple idea. Notice that the point $(7, 8/3)$ lies inside the parabola (easy, since $y(7)=0<8/3$) and it is also a point on the line $x=3y-1$. Hence, a point of the line lies inside the parabola which means that the line will intersect the parabola in two points (unless the line is vertical which is not the case here).

It's true that if the parabola opens upward and any point on the parabola is directly below any point on the line, then the parabola and line intersect in two distinct points. But unless you already have this fact available, it merits justification in its own right, and that's not going to be easier than proving the original fact (one method to prove the latter fact, for example, is using the discriminant).

In any case, it's not too much work to prove the claim in this question using the discriminant: We can rewrite the line as $y = \frac{1}{3} x + \frac{1}{3}$ and the parabola as $y = 3 x^2 - 17 x - 28$. Thus, the $x$-coordinates of the intersection points are the solutions of $$3 x^2 - \frac{52}{3} x - \frac{85}{3} = 0 .$$ Computing the discriminant of the l.h.s. requires some arithmetic, but we only need its sign: In our case, $a > 0, c < 0$, so $b^2 - 4 a c > 0$ and hence there are two distinct solutions.

Alternatively, applying Descartes' Rule of Signs tells us more: That there is exactly one positive solution and one negative solution.