3x^2-3x-33=0

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

3x^2-3x-33=0

a = 3; b = -3; c = -33;
Δ = b2-4ac
Δ = -32-4·3·(-33)
Δ = 405
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:

$\sqrt{\Delta}=\sqrt{405}=\sqrt{81*5}=\sqrt{81}*\sqrt{5}=9\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9\sqrt{5}}{2*3}=\frac{3-9\sqrt{5}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9\sqrt{5}}{2*3}=\frac{3+9\sqrt{5}}{6}
$