<s>[INST] Solving $TC = 7+2x+x^2$ and $TR = 10x$ Was wondering whether you might be able to take a look over my working and see if I'm doing this right! I was using the box method before, but have tried to do this one this way!

Find the break-even points in the case where total cost function $TC=7+2x+x^2$ and total revenue function is $TR=10x$.

Therefore, we need to solve: $$TC = TR$$
$$7+2x+x^2=10x\\
\therefore x^2+2x+7-10x=0\\
\therefore x^2-8x+7 = 0
$$
Now having the equation in the quadratic form $ax^2+bx+c$, we can solve using the quadratic formula:
$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\
$$
Plugging in the values $a=1$, $b=-8$ and $c=7$, we get:
$$\begin{align}
x & = \frac{-(-8)\pm \sqrt{(-8)^2-(4\cdot1\cdot7)}}{2\cdot1}\\\\
& = \frac{8\pm \sqrt{64-28}}{2}\\\\
& = \frac{8\pm \sqrt{36}}{2}\\\\
& = \frac{8\pm 6}{2}\\
\end{align}
$$
Therefore $x$ can equal:
$$\begin{align}\\
x & =\frac{8+6}{2}\\\\
& = \frac{14}{2}\\\\
& = 7
\end{align}
$$
OR:
$$\begin{align}\\
x & =\frac{8-6}{2}\\\\
& = \frac{2}{2}\\\\
& = 1
\end{align}
$$
Therefore, the break-even points could equal either $7$ or $1$.
 [/INST] Yes, your follow through is correct. Although I would like to point out one thing: You didn't have to use the quadratic formula to find the roots of the polynomial.
Whenever you get a polynomial, factoring usually is the quickest way to find the roots.
For your polynomial, $x^2-8x+7=0$, we see that we need two numbers to have a product of $7$ and sum to $8$. $1,7$ fit quite nicely. Thus, your quadratic can be factored into$$x^2-8x+7=(x-1)(x-7)=0\tag1$$
From which we see $x=1,7$ the two solutions.
Personally, I think it's quicker to factor than use the quadratic formula. Although the quadratic formula does have the nice feature of working on $any$ polynomial.
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