6c^2+20c+6=0

Solution for 6c^2+20c+6=0 equation:

6c^2+20c+6=0

a = 6; b = 20; c = +6;
Δ = b2-4ac
Δ = 202-4·6·6
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16}{2*6}=\frac{-36}{12}
=-3
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16}{2*6}=\frac{-4}{12}
=-1/3
$