32=(2x+8)(x+4)

Solution for 32=(2x+8)(x+4) equation:

32=(2x+8)(x+4)

We move all terms to the left:

32-((2x+8)(x+4))=0

We multiply parentheses ..

-((+2x^2+8x+8x+32))+32=0


We calculate terms in parentheses: -((+2x^2+8x+8x+32)), so:

(+2x^2+8x+8x+32)

We get rid of parentheses

2x^2+8x+8x+32

We add all the numbers together, and all the variables

2x^2+16x+32

Back to the equation:

-(2x^2+16x+32)


We get rid of parentheses

-2x^2-16x-32+32=0

We add all the numbers together, and all the variables

-2x^2-16x=0

a = -2; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·(-2)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*-2}=\frac{0}{-4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*-2}=\frac{32}{-4}
=-8
$