2x(x+7)=(x-4)(x+7)

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Solution for 2x(x+7)=(x-4)(x+7) equation:



2x(x+7)=(x-4)(x+7)
We move all terms to the left:
2x(x+7)-((x-4)(x+7))=0
We multiply parentheses
2x^2+14x-((x-4)(x+7))=0
We multiply parentheses ..
2x^2-((+x^2+7x-4x-28))+14x=0
We calculate terms in parentheses: -((+x^2+7x-4x-28)), so:
(+x^2+7x-4x-28)
We get rid of parentheses
x^2+7x-4x-28
We add all the numbers together, and all the variables
x^2+3x-28
Back to the equation:
-(x^2+3x-28)
We add all the numbers together, and all the variables
2x^2+14x-(x^2+3x-28)=0
We get rid of parentheses
2x^2-x^2+14x-3x+28=0
We add all the numbers together, and all the variables
x^2+11x+28=0
a = 1; b = 11; c = +28;
Δ = b2-4ac
Δ = 112-4·1·28
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-3}{2*1}=\frac{-14}{2} =-7 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+3}{2*1}=\frac{-8}{2} =-4 $

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