4800-560x+12x^2=0

Solution for 4800-560x+12x^2=0 equation:

4800-560x+12x^2=0

a = 12; b = -560; c = +4800;
Δ = b2-4ac
Δ = -5602-4·12·4800
Δ = 83200
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:

$\sqrt{\Delta}=\sqrt{83200}=\sqrt{6400*13}=\sqrt{6400}*\sqrt{13}=80\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-560)-80\sqrt{13}}{2*12}=\frac{560-80\sqrt{13}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-560)+80\sqrt{13}}{2*12}=\frac{560+80\sqrt{13}}{24}
$