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

Solution for (x-2)(x+4)=(-x-4)(x+2) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

(x-2)(x+4)-((-1x-4)(x+2))=0

We multiply parentheses ..

(+x^2+4x-2x-8)-((-1x-4)(x+2))=0


We calculate terms in parentheses: -((-1x-4)(x+2)), so:

(-1x-4)(x+2)

We multiply parentheses ..

(-1x^2-2x-4x-8)

We get rid of parentheses

-1x^2-2x-4x-8

We add all the numbers together, and all the variables

-1x^2-6x-8

Back to the equation:

-(-1x^2-6x-8)


We get rid of parentheses

x^2+1x^2+4x-2x+6x-8+8=0

We add all the numbers together, and all the variables

2x^2+8x=0

a = 2; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·2·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*2}=\frac{-16}{4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*2}=\frac{0}{4}
=0
$