0=3x^2+24x+6

Solution for 0=3x^2+24x+6 equation:

0=3x^2+24x+6

We move all terms to the left:

0-(3x^2+24x+6)=0

We add all the numbers together, and all the variables

-(3x^2+24x+6)=0

We get rid of parentheses

-3x^2-24x-6=0

a = -3; b = -24; c = -6;
Δ = b2-4ac
Δ = -242-4·(-3)·(-6)
Δ = 504
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-6\sqrt{14}}{2*-3}=\frac{24-6\sqrt{14}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+6\sqrt{14}}{2*-3}=\frac{24+6\sqrt{14}}{-6}
$