4(x+2)(x-2)=(2x)2+8x

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Solution for 4(x+2)(x-2)=(2x)2+8x equation:



4(x+2)(x-2)=(2x)2+8x
We move all terms to the left:
4(x+2)(x-2)-((2x)2+8x)=0
We add all the numbers together, and all the variables
-(+2x^2+8x)+4(x+2)(x-2)=0
We use the square of the difference formula
-(+2x^2+8x)+x^2-4=0
We get rid of parentheses
-2x^2+x^2-8x-4=0
We add all the numbers together, and all the variables
-1x^2-8x-4=0
a = -1; b = -8; c = -4;
Δ = b2-4ac
Δ = -82-4·(-1)·(-4)
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{3}}{2*-1}=\frac{8-4\sqrt{3}}{-2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{3}}{2*-1}=\frac{8+4\sqrt{3}}{-2} $

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