80x^2+120x+41=0

Solution for 80x^2+120x+41=0 equation:

80x^2+120x+41=0

a = 80; b = 120; c = +41;
Δ = b2-4ac
Δ = 1202-4·80·41
Δ = 1280
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:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-16\sqrt{5}}{2*80}=\frac{-120-16\sqrt{5}}{160}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+16\sqrt{5}}{2*80}=\frac{-120+16\sqrt{5}}{160}
$