192x^2-240x+50=0

Solution for 192x^2-240x+50=0 equation:

192x^2-240x+50=0

a = 192; b = -240; c = +50;
Δ = b2-4ac
Δ = -2402-4·192·50
Δ = 19200
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:

$\sqrt{\Delta}=\sqrt{19200}=\sqrt{6400*3}=\sqrt{6400}*\sqrt{3}=80\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-240)-80\sqrt{3}}{2*192}=\frac{240-80\sqrt{3}}{384}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-240)+80\sqrt{3}}{2*192}=\frac{240+80\sqrt{3}}{384}
$