2(x+1)(x-4)=4(x-2)(x+2)

Solution for 2(x+1)(x-4)=4(x-2)(x+2) equation:

2(x+1)(x-4)=4(x-2)(x+2)

We move all terms to the left:

2(x+1)(x-4)-(4(x-2)(x+2))=0

We use the square of the difference formula

x^2+2(x+1)(x-4)+4=0

We multiply parentheses ..

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

We multiply parentheses

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

We add all the numbers together, and all the variables

3x^2-6x-4=0

a = 3; b = -6; c = -4;
Δ = b2-4ac
Δ = -62-4·3·(-4)
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{21}}{2*3}=\frac{6-2\sqrt{21}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{21}}{2*3}=\frac{6+2\sqrt{21}}{6}
$