3x^2+12x-24=36

Solution for 3x^2+12x-24=36 equation:

3x^2+12x-24=36

We move all terms to the left:

3x^2+12x-24-(36)=0

We add all the numbers together, and all the variables

3x^2+12x-60=0

a = 3; b = 12; c = -60;
Δ = b2-4ac
Δ = 122-4·3·(-60)
Δ = 864
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:

$\sqrt{\Delta}=\sqrt{864}=\sqrt{144*6}=\sqrt{144}*\sqrt{6}=12\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{6}}{2*3}=\frac{-12-12\sqrt{6}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{6}}{2*3}=\frac{-12+12\sqrt{6}}{6}
$