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

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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 $

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