3c2+4c+8=-6c

Solution for 3c2+4c+8=-6c equation:

3c^2+4c+8=-6c

We move all terms to the left:

3c^2+4c+8-(-6c)=0

We get rid of parentheses

3c^2+4c+6c+8=0

We add all the numbers together, and all the variables

3c^2+10c+8=0

a = 3; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·3·8
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2}{2*3}=\frac{-12}{6}
=-2
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2}{2*3}=\frac{-8}{6}
=-1+1/3
$