<s>[INST] At what point does normal line intersect curve second time? 
At what point does the normal line to $y=-5+4x+3x^2$ at $(1,2)$ intersect the parabola a second time? 

$y'=6x+4$
$m_{tangent}=6(1)+4=10$
$m_{normal}=-\dfrac{1}{10}$  
$y=f'(1)(x-1)+f(1)$
$=(-1/10)(x-1)+2$
$y=-1/10+21/10 \implies$ equation of normal line  
Then, set normal line = curve to find point of intersection: 
$(-1/10x+21/10=-5+4x+3x^2)(-10) \rightarrow$ I multiplied by $-10$ to get rid of the $-1/10x$ fraction.  
$x=60-4x-30x^2$
$-5(6x^2+x-12)=0$  
Then I used the quadratic formula to find the $x$ values. I got $x= \dfrac{4}{3},-\dfrac{3}{2}$ and this is wrong. 
How do I find the $x$ values and $y$ values and what am I doing wrong? Thanks. 
 [/INST] Correcting some algebra, will look into the analytic geometry when I have time.
$$C_{norm}: y= \frac{21}{10}-\frac{x}{10}$$
Then you have $$\frac{21}{10}-\frac{x}{10}=-5+4x+3x^2 \iff -3x^2-\frac{41x}{10}+\frac{71}{10}=0 \iff -\frac{1}{10} ((x-1)(71+30x))=0 \iff (x-1)(71+30x)=0 \iff x=1 \; \mathrm{or} \; x=-\frac{71}{30}$$
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