12x^2-48x+16=0

Solution for 12x^2-48x+16=0 equation:

12x^2-48x+16=0

a = 12; b = -48; c = +16;
Δ = b2-4ac
Δ = -482-4·12·16
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16\sqrt{6}}{2*12}=\frac{48-16\sqrt{6}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16\sqrt{6}}{2*12}=\frac{48+16\sqrt{6}}{24}
$