3x^2+6x-12=27

Solution for 3x^2+6x-12=27 equation:

3x^2+6x-12=27

We move all terms to the left:

3x^2+6x-12-(27)=0

We add all the numbers together, and all the variables

3x^2+6x-39=0

a = 3; b = 6; c = -39;
Δ = b2-4ac
Δ = 62-4·3·(-39)
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{-(6)-6\sqrt{14}}{2*3}=\frac{-6-6\sqrt{14}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{14}}{2*3}=\frac{-6+6\sqrt{14}}{6}
$