(32-2x)*(24-2x)=568

Solution for (32-2x)*(24-2x)=568 equation:

(32-2x)(24-2x)=568

We move all terms to the left:

(32-2x)(24-2x)-(568)=0

We add all the numbers together, and all the variables

(-2x+32)(-2x+24)-568=0

We multiply parentheses ..

(+4x^2-48x-64x+768)-568=0

We get rid of parentheses

4x^2-48x-64x+768-568=0

We add all the numbers together, and all the variables

4x^2-112x+200=0

a = 4; b = -112; c = +200;
Δ = b2-4ac
Δ = -1122-4·4·200
Δ = 9344
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:

$\sqrt{\Delta}=\sqrt{9344}=\sqrt{64*146}=\sqrt{64}*\sqrt{146}=8\sqrt{146}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-8\sqrt{146}}{2*4}=\frac{112-8\sqrt{146}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+8\sqrt{146}}{2*4}=\frac{112+8\sqrt{146}}{8}
$