Question: Let $v:C\to \mathbb{R}^m$ be a $C^1$-vectorfield, where $C$ is a compact subset of $\mathbb{R}^m$. I want to prove that there is a constant $L>0$ such that $$\lvert Dv\_u \cdot w\rvert \leq L\cdot\lvert w \rvert$$ for all $u\in C$ and $w\in \mathbb{R}^m$, where $Dv\_u$ denotes the jacobian of $v$ at $u$. By using the unit sphere S=$\{w\in \mathbb{R}^m:\lvert x\rvert=1\}$, the linearity of $Dv\_u$ and the Minimum-Maximum-Theorem, I can show that for every fixed $u\in C$ there exists an $L>0$ such that the inequality holds. How do i proof that there is an $L>0$ for every $u\in \mathbb{R}^m$?

Answer: 
$\frac {c^2 - 14c - 28}{(c-7)^2} = -\frac {72}{5}$

I think the easiest thing to do is cross multiply.

$5(c^2 - 14c - 28) = -72(c-7)^2$

That looks like it is going to make some big numbers, but it might be nicer than it, at first looks. Expanding the right handd side:

$5(c^2 - 14c - 28) = -72(c^2 - 14c + 49)$

We have $c^2-14c$ on both sides of the equation, and we can combine those.

$77(c^2 - 14c) - 5\cdot 28 + 72\cdot 49 = 0$

We have some arithmetic to do, but notice that every term is divisible by 7.

$11(c^2 - 14c) - 5\cdot 4 + 72\cdot 7 = 0\\
11(c^2 - 14c) - 20 + 504 = 0\\
11(c^2 - 14c) + 484 = 0\\
(c^2 - 14c) + 44 = 0\\
$
this seems like a homework problem. Therefore, I'm happy to guide you, but you need to do some of the work yourself. Start out by expanding $(c - 7)^2$. This yields $c^2 - 14c + 49$. Then, cross-multiply to get rid of the denominators. That is, multiply both sides of the equation by $\frac{5}{5}$, and then multiply both sides by $\frac{c^2 - 14c + 49}{c^2 - 14c + 49}$. After cancelling, you should now have NO denominators. 

Since the result you are looking for is equal to zero, pick a side of the equation, and subtract all those terms from both sides. That should leave a zero on one side of the equation, and a nasty-looking formula on the other.

Now, collect like terms together. Look at your $c^2$, $c$, and constant terms, and see if you can combine some. This should give you a polynomial with three terms.

Now, see if you can simplify. Since one side is zero, you can freely divide by any constant, and that side will remain zero. Think about what constant you would need to divide by to get $c^2$ all by itself (since that is how it is in the equation you are trying to mimic).

Now you should be at your answer!