xsquared+2x-8=0 solve by factoring?

( x + 4 ) ( x - 2 ) = 0

x = - 4 , x = 2

x^2 + 2x - 8 = 0

This is a quadratic written in the classical format:
ax^2 + bx + c = 0

In your problem,
a = 1
b = 2
c = -8

You are looking for two factors
call them (x-m) and (x-m)

(x-m)(x-n) = x^2 - (m+n)x + mn
obviously you must find m and n such that:
-(m+n) = 2
mn = -8

The second one tells you that one must be positive and one must be negative.

The first one tells you that the absolute value of the negative one is greater than the absolute value of the positive one, by 2

Try
m = +2 and n=-4

then, (x-m) becomes (x-2)
and (x-n) becomes (x+4)

[yes, the signs can be tricky]

(x-2)(x+4) = x^2 -2x +4x -8 = x^2 + 2x - 8

We have our factors. Now, to solve the equation
-----

x^2 + 2x - 8 = 0
becomes
(x-2)(x+4) = 0

For a product to yield zero, one of the factors must be zero. You have no choice.

Therefore, there are two possibilities:
x-2 = 0 (therefore, x = +2)
and
x+4 = 0 (therefore, x = -4)

x² + 2x - 8 = 0
(x + 4)(x - 2) = 0
x + 4 = 0 ; x - 2 = 0
x = -4 ; x = 2

x^2 + 2x - 8 = 0
x^2 + 4x - 2x - 8 = 0
(x^2 + 4x) - (2x + 8) = 0
x(x + 4) - 2(x + 4) = 0
(x + 4)(x - 2) = 0

x + 4 = 0
x = -4

x - 2 = 0
x = 2

∴ x = -4, 2

Start out by putting ( ) ( ) first
than (x ) (x )
than you decide your signs (x- ) (x+ )
than what times what gives you 8 and subtracts to get you 2.
(x-2) (x+4)
there you go solved by factoring....

x^2+2x-8=0
(x + 4)(x - 2) = 0
x + 4 = 0 or x - 2 = 0
x = -4 or x = 2

(x+4)(x-2)=0

x = -4 or 2

x²+2x-8=0
(x+4)(x-2) = 0
x=-4; x=2



.

x^2 + 2x - 8 = 0
(x + 4)(x -2) = 0
x = -4, 2