3x^2+18x-33=12

Solution for 3x^2+18x-33=12 equation:

3x^2+18x-33=12

We move all terms to the left:

3x^2+18x-33-(12)=0

We add all the numbers together, and all the variables

3x^2+18x-45=0

a = 3; b = 18; c = -45;
Δ = b2-4ac
Δ = 182-4·3·(-45)
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{-(18)-12\sqrt{6}}{2*3}=\frac{-18-12\sqrt{6}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+12\sqrt{6}}{2*3}=\frac{-18+12\sqrt{6}}{6}
$