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

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
$