3u^2-12u+6=0

Solution for 3u^2-12u+6=0 equation:

3u^2-12u+6=0

a = 3; b = -12; c = +6;
Δ = b2-4ac
Δ = -122-4·3·6
Δ = 72
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{2}}{2*3}=\frac{12-6\sqrt{2}}{6}
$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{2}}{2*3}=\frac{12+6\sqrt{2}}{6}
$