3x2+12x+1=0

Solution for 3x2+12x+1=0 equation:

3x^2+12x+1=0

a = 3; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·3·1
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{33}}{2*3}=\frac{-12-2\sqrt{33}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{33}}{2*3}=\frac{-12+2\sqrt{33}}{6}
$