(32-2x)(24-2x)=128

Solution for (32-2x)(24-2x)=128 equation:

(32-2x)(24-2x)=128

We move all terms to the left:

(32-2x)(24-2x)-(128)=0

We add all the numbers together, and all the variables

(-2x+32)(-2x+24)-128=0

We multiply parentheses ..

(+4x^2-48x-64x+768)-128=0

We get rid of parentheses

4x^2-48x-64x+768-128=0

We add all the numbers together, and all the variables

4x^2-112x+640=0

a = 4; b = -112; c = +640;
Δ = b2-4ac
Δ = -1122-4·4·640
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-48}{2*4}=\frac{64}{8}
=8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+48}{2*4}=\frac{160}{8}
=20
$