Simultaneous equations - How does the following make sense The following example is derived from the video
https://www.youtube.com/watch?v=xee7Qqqd2eU ,
At the 5 minute mark the two corresponding lines of work are as follows:- 
$$2x^2-x-10 = 0$$ equal to $$(2x-5)(x+2) = 0 $$
However when I try to use my factoring technique to prove it, I cannot do it.
What I know is to get from this line $$2x^2-x-10 = 0$$ to the next I must find a value that multiplies to $-20$ (which is $2 \times -10$), and adds to $-1$.
I cant find these values, I am missing something because there are no issues completing this in the video.
Please help as I know it must be something simple and it is very irritating that I have spent this long on one GCSE question, I want to get it out the way ! 
 A: $$2x^2-x-10=(ax+b)(cx+d)$$
$$(ax+b) \cdot (cx+d)=a \cdot c \cdot x^2+a \cdot d \cdot x+b \cdot c \cdot x+b \cdot d$$
Where we could use the FOIL mnemonic. Where we multiply First Outer Inner Last. You can then compare coefficients to get,
$$2x^2-x-10=(2x-5)(x+2)$$
Alternatively just guess $a=2$ and $c=1$, and then deduce the other variables. That's what I did ;)
As noted by @Peter, You should also know that you can always pull out the quadratic formula to factor. You will have to multiply by an appropriate constant latter though! It seems like this method might defeat the purpose of factoring though.
A: The following simple explanation has got me to the point of understanding where I need to be :- 
$$2x^2 - x - 10 = 0$$
Find two values that multiply to -20 (2*-10) and add to -1.
$$4, -5$$
replace the -x with the newfound +4x and -5x, giving:
$$2x^2+4x-5x-10 = 0$$
Put into brackets to get rid of the x^2 term,
$$2x(x+2)   (x+2)$$
2x(x+2) corresponds to this term $$2x^2+4$$ 
and to make (x+2) correspond to 5x-10, multiply the second bracket by -5, giving 
$$2x(x+2)-5(x+2)$$
Now collect like terms $$(2x-5)(x+2)$$
and that is how he done it !
Still really appreciate everyone's help the comments definitely acted as stepping stones for me to get an understanding . 
A: You can either use the quadratic formula for a general quadratic $ax^2+bx+c=0$, which is
$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ 
but if you are not familiar with this I will say no more on it.
Or you can complete the square, or write
$$2x^2 \color{blue}{-x}-10=0$$
$$\implies 2x^2+\color{blue}{4x-5x}-10=0\tag{1}$$
$$\implies 2x(x+2)-5(x+2)=0\tag{2}$$
$$\implies \color{red}{(2x-5)(x+2)}\tag{3}=0$$
Notice, in the part marked $\color{blue}{\mathrm{blue}}$ I have simply rewritten $-x$ as $4x-5x$.
To get from $(1)$ to $(2)$ you factorize out the common factors of $2x$ and $-5$.
To get from $(2)$ to $(3)$ you factorize out the common factor $(x+2)$
Note that this method is only valid when you have two numbers whose product is $-20$ and sum is $-1$. 
Or, put in another way for the general quadratic equation $ax^2 +bx +c$, this inspection method is only valid iff you can find two numbers whose product is $ac$ and sum is $b$. Otherwise using the quadratic formula is the best method as it suits all scenarios.
For completing the square:
$$2x^2 -x-10=0$$
$$\implies 2\left(x^2 -\frac{x}{2}-\frac{3}{2}\right)=0$$
$$\implies 2\left(\left(x -\frac{1}{4}\right)^2-\frac{1}{16}-\frac{80}{16}\right)=0$$
$$\implies 2\left(\left(x -\frac{1}{4}\right)^2-\left(\frac{9}{4}\right)^2\right)=0\tag{A}$$
$$\implies 2\left(\left(x -\frac{1}{4}+\frac{9}{4}\right)\left(x -\frac{1}{4}-\frac{9}{4}\right)\right)\tag{B}=0$$
$$\implies 2\left((x+2)\left(x -\frac{5}{2}\right)\right)=0$$
$$\implies \color{red}{(2x-5)(x+2)}=0$$
which is the same as $(3)$ before. 
From $(\mathrm{A})$ to $(\mathrm{B})$ I used the difference of two squares formula such that $\alpha^2-\beta^2=(\alpha+\beta)(\alpha-\beta) \space \space\forall \space \alpha,\beta \in \mathbb{R}$
Hope this helps.  
