500+(x-500)x0.1=2,000

Solution for 500+(x-500)x0.1=2,000 equation:

500+(x-500)x0.1=2.000

We move all terms to the left:

500+(x-500)x0.1-(2.000)=0

We add all the numbers together, and all the variables

(x-500)x0.1+500-2=0

We add all the numbers together, and all the variables

(x-500)x0.1+498=0

We multiply parentheses

x^2-500x+498=0

a = 1; b = -500; c = +498;
Δ = b2-4ac
Δ = -5002-4·1·498
Δ = 248008
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:

$\sqrt{\Delta}=\sqrt{248008}=\sqrt{4*62002}=\sqrt{4}*\sqrt{62002}=2\sqrt{62002}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-500)-2\sqrt{62002}}{2*1}=\frac{500-2\sqrt{62002}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-500)+2\sqrt{62002}}{2*1}=\frac{500+2\sqrt{62002}}{2}
$