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

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

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

We move all terms to the left:

4x(x+2)-((x-2)(x-2))=0

We multiply parentheses

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+x^2-2x-2x+4)), so:

(+x^2-2x-2x+4)

We get rid of parentheses

x^2-2x-2x+4

We add all the numbers together, and all the variables

x^2-4x+4

Back to the equation:

-(x^2-4x+4)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3x^2+12x-4=0

a = 3; b = 12; c = -4;
Δ = b2-4ac
Δ = 122-4·3·(-4)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{3}}{2*3}=\frac{-12-8\sqrt{3}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{3}}{2*3}=\frac{-12+8\sqrt{3}}{6}
$