3x^2+6x-24=4

Solution for 3x^2+6x-24=4 equation:

3x^2+6x-24=4

We move all terms to the left:

3x^2+6x-24-(4)=0

We add all the numbers together, and all the variables

3x^2+6x-28=0

a = 3; b = 6; c = -28;
Δ = b2-4ac
Δ = 62-4·3·(-28)
Δ = 372
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:

$\sqrt{\Delta}=\sqrt{372}=\sqrt{4*93}=\sqrt{4}*\sqrt{93}=2\sqrt{93}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{93}}{2*3}=\frac{-6-2\sqrt{93}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{93}}{2*3}=\frac{-6+2\sqrt{93}}{6}
$