3x^2-200x+2500=0

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

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