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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x^2+12x=0

a = -3; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-3)·0
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{-(12)-12}{2*-3}=\frac{-24}{-6}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-3}=\frac{0}{-6}
=0
$