200x^2-80x+6=0

Solution for 200x^2-80x+6=0 equation:

200x^2-80x+6=0

a = 200; b = -80; c = +6;
Δ = b2-4ac
Δ = -802-4·200·6
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-40}{2*200}=\frac{40}{400}
=1/10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+40}{2*200}=\frac{120}{400}
=3/10
$