(20-2x)(32-2x)=448

Solution for (20-2x)(32-2x)=448 equation:

(20-2x)(32-2x)=448

We move all terms to the left:

(20-2x)(32-2x)-(448)=0

We add all the numbers together, and all the variables

(-2x+20)(-2x+32)-448=0

We multiply parentheses ..

(+4x^2-64x-40x+640)-448=0

We get rid of parentheses

4x^2-64x-40x+640-448=0

We add all the numbers together, and all the variables

4x^2-104x+192=0

a = 4; b = -104; c = +192;
Δ = b2-4ac
Δ = -1042-4·4·192
Δ = 7744
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7744}=88$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-104)-88}{2*4}=\frac{16}{8}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-104)+88}{2*4}=\frac{192}{8}
=24
$