3x^2-33x-162=0

Solution for 3x^2-33x-162=0 equation:

3x^2-33x-162=0

a = 3; b = -33; c = -162;
Δ = b2-4ac
Δ = -332-4·3·(-162)
Δ = 3033
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:

$\sqrt{\Delta}=\sqrt{3033}=\sqrt{9*337}=\sqrt{9}*\sqrt{337}=3\sqrt{337}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-3\sqrt{337}}{2*3}=\frac{33-3\sqrt{337}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+3\sqrt{337}}{2*3}=\frac{33+3\sqrt{337}}{6}
$