f^2+6f+6=0

Solution for f^2+6f+6=0 equation:

f^2+6f+6=0

a = 1; b = 6; c = +6;
Δ = b2-4ac
Δ = 62-4·1·6
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

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