Someone please explain to me how x1=2 and x2=6 in this equation:

y=x^2-8x+12

We are supposed to find their zeros(nullpoints) or whatever?

isn't it like this:
x^2-8x-12=0
x=4+- sq root of 16+12
x=4+- sq root of 28
etc??

where did I do wrong

Δ = (-8)² - 4 x 1 x 12
Δ = 64 - 48
Δ = 16

x = ( -(-8) ∓ √Δ ) / ( 2 x 1)
x = ( 8 ∓ 4 ) / 2
x = 2 or 6

Or since this one is factorable. Set y to 0 and solve for binomial pairs.

x^2 - 8x + 12 = 0

(x - 2)(x - 6) = 0

x = 2 or x = 6

Right but how do you get this?

(x - 2)(x - 6) = 0

That's where I lost it I think, I forgot how to correctly count this :P.
I have def done it before though for sure

It’s reverse FOILing. It sort of just something you have to be able to see in order to do it the short way (if the short way is available). If you can’t see the factors or they don’t exist in easy terms, feanur’s method always works for finding roots if they exist.

It’s two factors of the first term, x^2, multiplied with two factors of the third, 12. Since the coefficient for x^2 is 1, you only need to consider two factors of 12 which can add to get -8. In this case, -2 and -6.

If you were to multiply those two binomials out, you would use the FOIL method (First, Outer, Inner, Last) of distribution.

(x - 2)(x - 6) = x(x) + x(-6) + (-2)x + (-2)(-6)
= x^2 - 6x - 2x + 12
= x^2 - 8x + 12

This post was edited by timmayX on Mar 1 2018 11:49am

Give this a quick read, foundation for future mathematics if you're going further.

http://www.mathsisfun.com/algebra/factoring.html

KhanAcademy has an awesome video on factorizing quadratic equations. Let me see if I can find it for you.
/e It's not what I was trying to find, but he's got more videos apparently.

Meet Sal Khan, the man who teaches mastery of a subject rather than the basics. You'll find more videos in the link here, on the left of the page.

https://www.khanacademy.org/math/algebra/quadratics/solving-quadratic-equations-by-factoring/v/example-1-solving-a-quadratic-equation-by-factoring

This post was edited by Forg0tten on Mar 3 2018 07:15pm

He is amazing all the way through calc 3 + some diff eq stuff. Really incredible videos, great tools for learning

ok so another question
we have:
4x-5x^2=0

factorized to x(4-5x)=0 right?

Now how to solve x out of this?

It's supposed to be 2 x's here which are 0 and 4/5 (in decimal 0 and 0,8)

how to get these? that's where I fail once again..

This is the same as the reverse FOILing some above. It just looks different. Feanur’s method with the quadratic equation still applies where a = -5, b = 4, and c = 0.

Or you can think of them as pairs again:
4x - 5x^2 = 0

x(4 - 5x) = 0
(x)(4 - 5x) = 0

You can even write it as (x + 0)(4 - 5x) = 0, if it helps to have it in the same form of binomial pairs.

So, x = 0 makes this true because the factored out x would distribute through, x(4 - 5x) = 0 —> 0(4 - 5(0)) = 0 —> 0 = 0.

Then the other pair 4 - 5x = 0 —> 4 = 5x —> x = 4/5.

Once you have them in factored binomial pairs, either one of them being equal to 0 will make the equation valid. So you set each one equal to 0 to find the x which makes each portion valid.