(20-2x)(12-2x)=1

Solution for (20-2x)(12-2x)=1 equation:

(20-2x)(12-2x)=1

We move all terms to the left:

(20-2x)(12-2x)-(1)=0

We add all the numbers together, and all the variables

(-2x+20)(-2x+12)-1=0

We multiply parentheses ..

(+4x^2-24x-40x+240)-1=0

We get rid of parentheses

4x^2-24x-40x+240-1=0

We add all the numbers together, and all the variables

4x^2-64x+239=0

a = 4; b = -64; c = +239;
Δ = b2-4ac
Δ = -642-4·4·239
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{-(-64)-4\sqrt{17}}{2*4}=\frac{64-4\sqrt{17}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+4\sqrt{17}}{2*4}=\frac{64+4\sqrt{17}}{8}
$