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?


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.
