2000+100x-0.1x^2=0

Solution for 2000+100x-0.1x^2=0 equation:

2000+100x-0.1x^2=0

a = -0.1; b = 100; c = +2000;
Δ = b2-4ac
Δ = 1002-4·(-0.1)·2000
Δ = 10800
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:

$\sqrt{\Delta}=\sqrt{10800}=\sqrt{3600*3}=\sqrt{3600}*\sqrt{3}=60\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-60\sqrt{3}}{2*-0.1}=\frac{-100-60\sqrt{3}}{-0.2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+60\sqrt{3}}{2*-0.1}=\frac{-100+60\sqrt{3}}{-0.2}
$