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 tagged calculus so with derivatives: the slope of a tangent to the given function is
$$y'=6x+7\implies\;\text{for any point on the graph }\;\;(a, 3a^2+7a-2)$$
the tangent line to the function at that point is
$$y-(3a^2+7a-2)=(6a+7)(x-a)\implies y=(6a+7)x-3a^2-2$$
and thus the $\;y\,-$ intercept is $\;-3a^2-2\;$ , and this is $\;10\;$ iff
$$-3a^2-2=10\iff a^2=-4$$
and this last equality can't be true in the real numbers
