3x^2+24x=6

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

3x^2+24x=6

We move all terms to the left:

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

a = 3; b = 24; c = -6;
Δ = b2-4ac
Δ = 242-4·3·(-6)
Δ = 648
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:

$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-18\sqrt{2}}{2*3}=\frac{-24-18\sqrt{2}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+18\sqrt{2}}{2*3}=\frac{-24+18\sqrt{2}}{6}
$