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

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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} $

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