x=3000+20x-0.1x^2

Solution for x=3000+20x-0.1x^2 equation:

x=3000+20x-0.1x^2

We move all terms to the left:

x-(3000+20x-0.1x^2)=0

We get rid of parentheses

0.1x^2-20x+x-3000=0

We add all the numbers together, and all the variables

0.1x^2-19x-3000=0

a = 0.1; b = -19; c = -3000;
Δ = b2-4ac
Δ = -192-4·0.1·(-3000)
Δ = 1561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{1561}}{2*0.1}=\frac{19-\sqrt{1561}}{0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{1561}}{2*0.1}=\frac{19+\sqrt{1561}}{0.2}
$