X^2-8x-20=0? factoring

When factoring x^2-8x-20=0 I know the factored answer is (x+2)(x-10)=0 I know that's the factored answer but where are they getting the +2 and the -10 from the orginal equation or another example
X^2-3x-4=0 the factored answer is (x-4)(x+1)=0 but again where did they get the -4 and the +1 please explain. Ps I can't use a calculator for the test if I could I would just graph it but out teacher wants to see the actual factoring process please help

(x + 2)(x - 10) = 0
x = -2 , 10

Edit: OK,
x^2 - 8x - 20 = 0 => find two numbers that add to -8 , multiply to -20? -10 & 2?
x^2 + 2x - 10x + 20 = 0
x(x + 2) - 10(x + 2) = 0
(x + 2)(x - 10) = 0

Now you can do the 2nd one right?

So you have (x+2)(x-10)=0

What you're doing is solving for the x value. So we'll solve for the first factor, (x+2).

Divide both sides by (x-10) and you get

(x+2)=0 or x+2=0

Then simply subtract the 2 from both sides to isolate the x and you get x= -2

Now solve for the other factor, (x-10).

Divide both sides by (x+2) and you get

(x-10)=0 or x-10=0

Then add 10 to both sides to isolate the x and you get x=10

So, the solutions are x= -2 or x= 10

Plug both of those values into the original equation to check if they are a solution.

(-2)^2-8(2)-20=0
4-16-20=0
20-20=0

x= -2 is a solution for the equation

(10)^2-8(10)-20=0
100-80-20=0
20-20=0

x=10 is also a solution for the equation.

You have to use the diamond-and-box method or the quadratic equation. Look those up via the soure BELOW and it'll teach you how to do that. Once you learn this, the reason for the numbers being the values that they are will be clear.

-4 times 1 is -4 this is a*c and -4+(-1) = -3 which is b.

ax^2 + bx + c = 0, they are sing ac method.

x^2-8x-20=0
(x+2)(x-10)=0

(x+2)=0 ==> x=-2
(x-10)=0==>x=10

X^2-3x-4=0
(x-4)(x+1)=0

(x-4)=0==>x=4
(x+1)=0==>x=-1