3c^2-12c+8=0

Solution for 3c^2-12c+8=0 equation:

3c^2-12c+8=0

a = 3; b = -12; c = +8;
Δ = b2-4ac
Δ = -122-4·3·8
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{3}}{2*3}=\frac{12-4\sqrt{3}}{6}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{3}}{2*3}=\frac{12+4\sqrt{3}}{6}
$