12x^2-320x+1599=0

Solution for 12x^2-320x+1599=0 equation:

12x^2-320x+1599=0

a = 12; b = -320; c = +1599;
Δ = b2-4ac
Δ = -3202-4·12·1599
Δ = 25648
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:

$\sqrt{\Delta}=\sqrt{25648}=\sqrt{16*1603}=\sqrt{16}*\sqrt{1603}=4\sqrt{1603}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-320)-4\sqrt{1603}}{2*12}=\frac{320-4\sqrt{1603}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-320)+4\sqrt{1603}}{2*12}=\frac{320+4\sqrt{1603}}{24}
$