<s>[INST] Solving for $p$ and $k$. For the question,
$$F(x)= x^{2} + 6x + 20 + k(x^{2} -3x -12)$$
I have to find the value of k and the value of $p$ given that the minimum value of $F(x)$ is $p$ and $F(-2) = p$.
What I did is: I expanded the expression of $F(x)$ and reached a term for $p= 12 - 2p$. I tried to complete the square of the two quadratics of $F(x)$ separately then added the two terms while also correcting for coefficients. I doubt this operation however went for it anyway due to any alternatives seem wrong.
I reached the term $-\frac {39}4k + 11$ then promptly equated it to $12 - 2k$.
I reached the value of $-\frac{4}{31}$ for $K$. 
My textbook has answers for $k = \frac{2}{7}$ and $p=\frac{80}{7}$.
 [/INST] A quadratic $ax^2+bx+c$ with a positive leading term $a\gt0$ attains its minimum at $-\cfrac{b}{2a}\,$.
Rewriting $F(x) = (1+k)x^2+(6-3k)x+20-12k\,$ it follows that: $$-\cfrac{6-3k}{2(1+k)}=-2 \quad \implies \quad k = \cfrac{2}{7}$$
(Since the leading term $1+k = 1 + \cfrac{2}{7}\gt 0$ this is a valid solution.)
Substituting $k$ back gives  $F(x)=\cfrac{9}{7} x^2+\cfrac{36}{7} x + \cfrac{116}{7}\,$, then $p=F(-2)=\cfrac{80}{7}\,$.
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