3a^2+6a+1=0

Solution for 3a^2+6a+1=0 equation:

3a^2+6a+1=0

a = 3; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·3·1
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{6}}{2*3}=\frac{-6-2\sqrt{6}}{6}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{6}}{2*3}=\frac{-6+2\sqrt{6}}{6}
$