3x^2-33x-9=0

Solution for 3x^2-33x-9=0 equation:

3x^2-33x-9=0

a = 3; b = -33; c = -9;
Δ = b2-4ac
Δ = -332-4·3·(-9)
Δ = 1197
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:

$\sqrt{\Delta}=\sqrt{1197}=\sqrt{9*133}=\sqrt{9}*\sqrt{133}=3\sqrt{133}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-3\sqrt{133}}{2*3}=\frac{33-3\sqrt{133}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+3\sqrt{133}}{2*3}=\frac{33+3\sqrt{133}}{6}
$