(8-2x)(6-2x)=16

Solution for (8-2x)(6-2x)=16 equation:

(8-2x)(6-2x)=16

We move all terms to the left:

(8-2x)(6-2x)-(16)=0

We add all the numbers together, and all the variables

(-2x+8)(-2x+6)-16=0

We multiply parentheses ..

(+4x^2-12x-16x+48)-16=0

We get rid of parentheses

4x^2-12x-16x+48-16=0

We add all the numbers together, and all the variables

4x^2-28x+32=0

a = 4; b = -28; c = +32;
Δ = b2-4ac
Δ = -282-4·4·32
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{17}}{2*4}=\frac{28-4\sqrt{17}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{17}}{2*4}=\frac{28+4\sqrt{17}}{8}
$