0=(20-2x)(16-2x)

Solution for 0=(20-2x)(16-2x) equation:

0=(20-2x)(16-2x)

We move all terms to the left:

0-((20-2x)(16-2x))=0

We add all the numbers together, and all the variables

-((-2x+20)(-2x+16))+0=0

We add all the numbers together, and all the variables

-((-2x+20)(-2x+16))=0

We multiply parentheses ..

-((+4x^2-32x-40x+320))=0


We calculate terms in parentheses: -((+4x^2-32x-40x+320)), so:

(+4x^2-32x-40x+320)

We get rid of parentheses

4x^2-32x-40x+320

We add all the numbers together, and all the variables

4x^2-72x+320

Back to the equation:

-(4x^2-72x+320)


We get rid of parentheses

-4x^2+72x-320=0

a = -4; b = 72; c = -320;
Δ = b2-4ac
Δ = 722-4·(-4)·(-320)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-8}{2*-4}=\frac{-80}{-8}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+8}{2*-4}=\frac{-64}{-8}
=+8
$