2x^2-160x+2000=0

Solution for 2x^2-160x+2000=0 equation:

2x^2-160x+2000=0

a = 2; b = -160; c = +2000;
Δ = b2-4ac
Δ = -1602-4·2·2000
Δ = 9600
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:

$\sqrt{\Delta}=\sqrt{9600}=\sqrt{1600*6}=\sqrt{1600}*\sqrt{6}=40\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-40\sqrt{6}}{2*2}=\frac{160-40\sqrt{6}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+40\sqrt{6}}{2*2}=\frac{160+40\sqrt{6}}{4}
$