6(x-2)+x(2-x)=(x-2)(6+x)

Solution for 6(x-2)+x(2-x)=(x-2)(6+x) equation:

6(x-2)+x(2-x)=(x-2)(6+x)

We move all terms to the left:

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

We add all the numbers together, and all the variables

6(x-2)+x(-1x+2)-((x-2)(x+6))=0

We multiply parentheses

-1x^2+6x+2x-((x-2)(x+6))-12=0

We multiply parentheses ..

-1x^2-((+x^2+6x-2x-12))+6x+2x-12=0


We calculate terms in parentheses: -((+x^2+6x-2x-12)), so:

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

We get rid of parentheses

x^2+6x-2x-12

We add all the numbers together, and all the variables

x^2+4x-12

Back to the equation:

-(x^2+4x-12)


We add all the numbers together, and all the variables

-1x^2+8x-(x^2+4x-12)-12=0

We get rid of parentheses

-1x^2-x^2+8x-4x+12-12=0

We add all the numbers together, and all the variables

-2x^2+4x=0

a = -2; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-2)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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