(16-2x)(12-2x)=64

Solution for (16-2x)(12-2x)=64 equation:

(16-2x)(12-2x)=64

We move all terms to the left:

(16-2x)(12-2x)-(64)=0

We add all the numbers together, and all the variables

(-2x+16)(-2x+12)-64=0

We multiply parentheses ..

(+4x^2-24x-32x+192)-64=0

We get rid of parentheses

4x^2-24x-32x+192-64=0

We add all the numbers together, and all the variables

4x^2-56x+128=0

a = 4; b = -56; c = +128;
Δ = b2-4ac
Δ = -562-4·4·128
Δ = 1088
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1088}=\sqrt{64*17}=\sqrt{64}*\sqrt{17}=8\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-8\sqrt{17}}{2*4}=\frac{56-8\sqrt{17}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+8\sqrt{17}}{2*4}=\frac{56+8\sqrt{17}}{8}
$