800=(60-2x)(40-2x)

Solution for 800=(60-2x)(40-2x) equation:

800=(60-2x)(40-2x)

We move all terms to the left:

800-((60-2x)(40-2x))=0

We add all the numbers together, and all the variables

-((-2x+60)(-2x+40))+800=0

We multiply parentheses ..

-((+4x^2-80x-120x+2400))+800=0


We calculate terms in parentheses: -((+4x^2-80x-120x+2400)), so:

(+4x^2-80x-120x+2400)

We get rid of parentheses

4x^2-80x-120x+2400

We add all the numbers together, and all the variables

4x^2-200x+2400

Back to the equation:

-(4x^2-200x+2400)


We get rid of parentheses

-4x^2+200x-2400+800=0

We add all the numbers together, and all the variables

-4x^2+200x-1600=0

a = -4; b = 200; c = -1600;
Δ = b2-4ac
Δ = 2002-4·(-4)·(-1600)
Δ = 14400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{14400}=120$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-120}{2*-4}=\frac{-320}{-8}
=+40
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+120}{2*-4}=\frac{-80}{-8}
=+10
$