5c2+12c+6=0

Solution for 5c2+12c+6=0 equation:

5c^2+12c+6=0

a = 5; b = 12; c = +6;
Δ = b2-4ac
Δ = 122-4·5·6
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{6}}{2*5}=\frac{-12-2\sqrt{6}}{10}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{6}}{2*5}=\frac{-12+2\sqrt{6}}{10}
$