0=3x^2-200x+2500

Solution for 0=3x^2-200x+2500 equation:

0=3x^2-200x+2500

We move all terms to the left:

0-(3x^2-200x+2500)=0

We add all the numbers together, and all the variables

-(3x^2-200x+2500)=0

We get rid of parentheses

-3x^2+200x-2500=0

a = -3; b = 200; c = -2500;
Δ = b2-4ac
Δ = 2002-4·(-3)·(-2500)
Δ = 10000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10000}=100$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-100}{2*-3}=\frac{-300}{-6}
=+50
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+100}{2*-3}=\frac{-100}{-6}
=16+2/3
$