2x^2-80x+400=0

Solution for 2x^2-80x+400=0 equation:

2x^2-80x+400=0

a = 2; b = -80; c = +400;
Δ = b2-4ac
Δ = -802-4·2·400
Δ = 3200
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:

$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-40\sqrt{2}}{2*2}=\frac{80-40\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+40\sqrt{2}}{2*2}=\frac{80+40\sqrt{2}}{4}
$