(12x^2)-48x+39=0

Solution for (12x^2)-48x+39=0 equation:

(12x^2)-48x+39=0

a = 12; b = -48; c = +39;
Δ = b2-4ac
Δ = -482-4·12·39
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-12\sqrt{3}}{2*12}=\frac{48-12\sqrt{3}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+12\sqrt{3}}{2*12}=\frac{48+12\sqrt{3}}{24}
$