100x-(2500+10x+0.2x^2)=0

Solution for 100x-(2500+10x+0.2x^2)=0 equation:

100x-(2500+10x+0.2x^2)=0

We get rid of parentheses

-0.2x^2-10x+100x-2500=0

We add all the numbers together, and all the variables

-0.2x^2+90x-2500=0

a = -0.2; b = 90; c = -2500;
Δ = b2-4ac
Δ = 902-4·(-0.2)·(-2500)
Δ = 6100
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:

$\sqrt{\Delta}=\sqrt{6100}=\sqrt{100*61}=\sqrt{100}*\sqrt{61}=10\sqrt{61}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(90)-10\sqrt{61}}{2*-0.2}=\frac{-90-10\sqrt{61}}{-0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(90)+10\sqrt{61}}{2*-0.2}=\frac{-90+10\sqrt{61}}{-0.4}
$