Show that no line with a y-int of 10 will ever be tangential to the curve with $y=3x^2+7x-2$ Show that no line with a y-int of 10 will ever be tangential to the curve with $y=3x^2+7x-2$.
Having trouble in showing this. So far these are my process. 


*

*Let line be $y=mx+10$

*$mx+10 = 3x^2+7x-2$

*$3x^2+(7-m)x-12=0$

*Apply quadratic formula

*$\frac{(7-m\pm\sqrt{(m^2-14m+193)}}6$


A bit stuck here. Maybe I've missed the whole point and complicated this. Any help is appreciated! Thanks in advance :)

You got (almost)
$$x_{1,2}={m-7\pm\sqrt{(m-7)^2+144}\over6}\ .$$
As $(m-7)^2+144>0$ for all $m$ this says that  you obtain two different points of intersection whatever the chosen slope $m$ is.
If the discriminant would have been, e.g., $m^2-5m+6$ then there would be two "special" values of $m$ for which the discriminant is $0$. The two lines with these slopes would then be tangents to the parabola.
