2x(6-2x)-4=4x(6-2x)+4

Solution for 2x(6-2x)-4=4x(6-2x)+4 equation:

2x(6-2x)-4=4x(6-2x)+4

We move all terms to the left:

2x(6-2x)-4-(4x(6-2x)+4)=0

We add all the numbers together, and all the variables

2x(-2x+6)-(4x(-2x+6)+4)-4=0

We multiply parentheses

-4x^2+12x-(4x(-2x+6)+4)-4=0


We calculate terms in parentheses: -(4x(-2x+6)+4), so:

4x(-2x+6)+4

We multiply parentheses

-8x^2+24x+4

Back to the equation:

-(-8x^2+24x+4)


We get rid of parentheses

-4x^2+8x^2-24x+12x-4-4=0

We add all the numbers together, and all the variables

4x^2-12x-8=0

a = 4; b = -12; c = -8;
Δ = b2-4ac
Δ = -122-4·4·(-8)
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{17}}{2*4}=\frac{12-4\sqrt{17}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{17}}{2*4}=\frac{12+4\sqrt{17}}{8}
$