6x2+12x+4=0

Solution for 6x2+12x+4=0 equation:

6x^2+12x+4=0

a = 6; b = 12; c = +4;
Δ = b2-4ac
Δ = 122-4·6·4
Δ = 48
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:

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