<s>[INST] Could translate/explain this for me? I have this problem: 
$$
  10x^2 - 7x - 12 = 0
$$
And apparently the method to factoring it is to find two numbers whose product is the same as the product of the coefficient of $x^2$ and the constant term, and whose sum is the same as the coefficient of $x$.
However, I can barely understand what that even means, and as far as I'm concerned it seems impossible with the expression given.
Could someone possibly just walk me through this method?
 [/INST] 
to find two numbers whose product is the same as the product of the coefficient of $x^2$ and the constant term, and whose sum is the same as the coefficient of $x$.

Take the smallest, innermost noun phrases first:

to find two numbers whose product is the same as the product of the coefficient of $x^2$ and the constant term, and whose sum is the same as the coefficient of $x$.

Identify those pieces in the given polynomial:
$$ \underbrace{10}_{\text{coefficient of $x^2$}}x^2 + \overbrace{(-7)}^{\text{coefficient of $x$}}x + \underbrace{(-12)}_{\text{constant term}} $$
Plug those in to the question, to simplify it:

to find two numbers whose product is the same as the product of $10$ and $-12$, and whose sum is the same as $-7$.

"The product of $10$ and $-12$" is $-120$.  Let's plug that in.

to find two numbers whose product is the same as $-120$, and whose sum is the same as $-7$.

Now the words "the same as" seem pointless, so let's get rid of them:

to find two numbers whose product is $-120$, and whose sum is $-7$.

An obvious way to proceed here is just to list all the pairs of numbers whose product is $-120$, and to see which of them have sum $-7$.  (We can list them all the ones where both numbers are integers, at least... hopefully that'll be enough.)  Now, $-120$ has lots of factors, so there's lots of possibilities to consider.  Any way of splitting up the factors on the RHS of
$$ -120 = -1\cdot 2\cdot 2 \cdot 2 \cdot 3\cdot 5 $$
into two groups will make a factorization of $-120$ which might work.  For example, we might try
$$ -1\cdot 2\cdot 3 \qquad\text{and}\qquad 2\cdot 2\cdot 5 $$
Those are $-6$ and $20$; their product is $-120$, but their sum is $14$, not $-7$ as we wished.
By playing around for a while, or by systematically checking every possibility, you'll eventually find that
$$ 2\cdot 2\cdot 2 \qquad\text{and}\qquad -1\cdot 3\cdot 5 $$
that is, $8$ and $-15$, have product $-120$ and sum $-7$.
The rest of the method, which you didn't mention, is to break up the $x$ term using these two numbers,
\begin{align*}
10x^2 - 7x - 12
&= 10x^2 - 15x + 8x - 12
\end{align*}
then to take out common factors from the left two terms and from the right two terms,
\begin{align*}
10x^2 - 7x - 12
&= 10x^2 - 15x + 8x - 12 \\
&= 5x(2x - 3) + 4(2x - 3)
\end{align*}
and then, if we did everything right, we'll find, as we did here, that there's now a common factor between the left and the right, which we take out:
\begin{align*}
10x^2 - 7x - 12
&= 10x^2 - 15x + 8x - 12 \\
&= 5x(2x - 3) + 4(2x - 3) \\
&= (5x+4)(2x-3)
\end{align*}
Factorization complete.
</s>