.2x^2-80x+2000=0

Solution for .2x^2-80x+2000=0 equation:

.2x^2-80x+2000=0

a = .2; b = -80; c = +2000;
Δ = b2-4ac
Δ = -802-4·.2·2000
Δ = 4800
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:

$\sqrt{\Delta}=\sqrt{4800}=\sqrt{1600*3}=\sqrt{1600}*\sqrt{3}=40\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-40\sqrt{3}}{2*.2}=\frac{80-40\sqrt{3}}{0.4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+40\sqrt{3}}{2*.2}=\frac{80+40\sqrt{3}}{0.4}
$