80x^2-160x+71=0

Solution for 80x^2-160x+71=0 equation:

80x^2-160x+71=0

a = 80; b = -160; c = +71;
Δ = b2-4ac
Δ = -1602-4·80·71
Δ = 2880
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:

$\sqrt{\Delta}=\sqrt{2880}=\sqrt{576*5}=\sqrt{576}*\sqrt{5}=24\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-24\sqrt{5}}{2*80}=\frac{160-24\sqrt{5}}{160}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+24\sqrt{5}}{2*80}=\frac{160+24\sqrt{5}}{160}
$