(2x+4)(4-x)=2(x^2-8x+16)

Solution for (2x+4)(4-x)=2(x^2-8x+16) equation:

(2x+4)(4-x)=2(x^2-8x+16)

We move all terms to the left:

(2x+4)(4-x)-(2(x^2-8x+16))=0

We add all the numbers together, and all the variables

(2x+4)(-1x+4)-(2(x^2-8x+16))=0

We multiply parentheses ..

(-2x^2+8x-4x+16)-(2(x^2-8x+16))=0


We calculate terms in parentheses: -(2(x^2-8x+16)), so:

2(x^2-8x+16)

We multiply parentheses

2x^2-16x+32

Back to the equation:

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


We get rid of parentheses

-2x^2-2x^2+8x-4x+16x+16-32=0

We add all the numbers together, and all the variables

-4x^2+20x-16=0

a = -4; b = 20; c = -16;
Δ = b2-4ac
Δ = 202-4·(-4)·(-16)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12}{2*-4}=\frac{-32}{-8}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12}{2*-4}=\frac{-8}{-8}
=1
$