3a^2+24a+4=0

Solution for 3a^2+24a+4=0 equation:

3a^2+24a+4=0

a = 3; b = 24; c = +4;
Δ = b2-4ac
Δ = 242-4·3·4
Δ = 528
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{33}}{2*3}=\frac{-24-4\sqrt{33}}{6}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{33}}{2*3}=\frac{-24+4\sqrt{33}}{6}
$