(1008)=(2x-4)(x+8)

Solution for (1008)=(2x-4)(x+8) equation:

(1008)=(2x-4)(x+8)

We move all terms to the left:

(1008)-((2x-4)(x+8))=0

We multiply parentheses ..

-((+2x^2+16x-4x-32))+1008=0


We calculate terms in parentheses: -((+2x^2+16x-4x-32)), so:

(+2x^2+16x-4x-32)

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2+12x-32

Back to the equation:

-(2x^2+12x-32)


We get rid of parentheses

-2x^2-12x+32+1008=0

We add all the numbers together, and all the variables

-2x^2-12x+1040=0

a = -2; b = -12; c = +1040;
Δ = b2-4ac
Δ = -122-4·(-2)·1040
Δ = 8464
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8464}=92$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-92}{2*-2}=\frac{-80}{-4}
=+20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+92}{2*-2}=\frac{104}{-4}
=-26
$