3x^2-33-18x=0

Solution for 3x^2-33-18x=0 equation:

3x^2-33-18x=0

a = 3; b = -18; c = -33;
Δ = b2-4ac
Δ = -182-4·3·(-33)
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-12\sqrt{5}}{2*3}=\frac{18-12\sqrt{5}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+12\sqrt{5}}{2*3}=\frac{18+12\sqrt{5}}{6}
$