(500-2x)(400-2x)=100000

Solution for (500-2x)(400-2x)=100000 equation:

(500-2x)(400-2x)=100000

We move all terms to the left:

(500-2x)(400-2x)-(100000)=0

We add all the numbers together, and all the variables

(-2x+500)(-2x+400)-100000=0

We multiply parentheses ..

(+4x^2-800x-1000x+200000)-100000=0

We get rid of parentheses

4x^2-800x-1000x+200000-100000=0

We add all the numbers together, and all the variables

4x^2-1800x+100000=0

a = 4; b = -1800; c = +100000;
Δ = b2-4ac
Δ = -18002-4·4·100000
Δ = 1640000
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:

$\sqrt{\Delta}=\sqrt{1640000}=\sqrt{40000*41}=\sqrt{40000}*\sqrt{41}=200\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1800)-200\sqrt{41}}{2*4}=\frac{1800-200\sqrt{41}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1800)+200\sqrt{41}}{2*4}=\frac{1800+200\sqrt{41}}{8}
$