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
If you assume that the line is tangent to the parabola then
$$
mx+10=3x^2+7x-2
$$
at some point. Simplifying the previous equation we have
$$
3x^2+(7-m)x-12=0.
$$
Since we are assuming they are tangential, the discriminant of the last equation should be equal to zero. That is,
$$
m^2-14m+193=0.
$$
Solving for $m$ using the quadratic formula you get 
$x=7+12i$ or $x=7-12i$, which are not solutions in the real numbers. Therefore the line and the parabola are not tangential.