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

Solution for 0=4(x+2)(x-4) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -(4(+x^2-4x+2x-8)), so:

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

We multiply parentheses

4x^2-16x+8x-32

We add all the numbers together, and all the variables

4x^2-8x-32

Back to the equation:

-(4x^2-8x-32)


We get rid of parentheses

-4x^2+8x+32=0

a = -4; b = 8; c = +32;
Δ = b2-4ac
Δ = 82-4·(-4)·32
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-24}{2*-4}=\frac{-32}{-8}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+24}{2*-4}=\frac{16}{-8}
=-2
$