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?
 A: 
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.
A: You have,
$$10x^{2}-7x-12=0$$
To factor this, I would use the quadratic formula,
that is,
$$x=\frac{7 \pm \sqrt{49-4(10)(-12)}}{20}$$
which solving leads to
$$x= \frac{-4}{5}$$ and $$x= \frac{3}{2}$$
to put this in the factored form now is easy as we can just write it as ,
$$(2x-3)(5x+4)$$ (notice the roots don't change, but I multiplied to get whole numbers)
You are also asking about a certain method.
If we rewrite your quadratic in an equivalent form
ie, 
$$x^2-\frac{7x}{10}-\frac{6}{5}=0$$
Then we can use the method that is to find the two numbers that multiply to -6/5 and add to -7/10.
Do you notice anything similar from our numbers?
A: Suppose you start with the expression $$(ax+b)(cx+d)=acx^2+(ad+bc)x+bd=px^2+qx+r$$
This helps us to see how the factoriastion we want relates to the polynomial we have. We know $p,q,r$ but we don't know $a,b,c,d$.
What we do know, though, is that if $a, b, c, d$ exist then comparing coefficients we get $abcd=pr$ and $ad+bc=q$. The two numbers $ad$ and $bc$ have sum $q$ and product $pr$.
Let's suppose we have any two numbers with $de=pr, e=\frac {pr}d$ (assuming $d\neq 0$, which is a trivial case) and $d+e=q$.
We then can write $$px^2+qx+r=px^2+(d+e)x+r=(px^2+dx)+(\frac {pr}dx+r)=x(px+d)+\frac rd(px+d)=$$$$=\left (x+\frac rd\right)(px+d)=\frac 1d(dx+r)(px+d)$$
Now if it matters that we are working with integers, note that $d\mid pr$ by construction, so $d$ can be cancelled.
A: The answer is: $(2x-3)(5x+4)$. You need to do some guessing that comes with experience. In this case, you assume whole numbers for the coefficients and look at factoring - $10 = 5\cdot2$ and $12 = 3\cdot4$, and you use this a as a first attempt.
A: To answer your question: Your method may be referring to Vieta's formulas.
First, if it is possible to factor the quadratic, there always exist such (real) numbers $x_1,x_2$:  
$$ax^2+bx+c=a(x-x_1)(x-x_2),\ \ a\neq 0$$
Finding the $x_1,x_2$ is your task when you are asked to factor. Then (use distributive property):
$$a(x-x_1)(x-x_2)=ax^2+x(-ax_1-ax_2)+ax_1x_2$$
$$-ax_1-ax_2=b,\ \ \ \ \ \ ax_1x_2=c$$
$$x_1+x_2=-\frac{b}{a},\ \ \ \ \ \ x_1x_2=\frac{c}{a}$$
In your case, $10x^2-7x-12=10(x-x_1)(x-x_2)$ with $$x_1+x_2=-\frac{-7}{10},\ \ \ \ \ \ x_1x_2=\frac{-12}{10}$$
$$x_1+x_2=0.7,\ \ \ \ \ \ x_1x_2=-1.2$$
$x_1=1.5,\, x_2=-0.8$ works by guessing (you can solve the system of course, but at this point you would use quadratic formula instead and it would miss the point of the method).  

In this case the method didn't work well. It can however work quite well when $a=1$ and the coefficients are integers.
E.g., see $x^2-4x-12=(x-x_1)(x-x_2)$. Then $$x_1+x_2=4,\ \ \ \ \ \ x_1x_2=-12$$
Guessing and checking can fairly easily get you $\{x_1,x_2\}=\{6, -2\}$, and so $$x^2-4x-12=(x-6)(x+2)$$
