3(2x^2-5x-33)=0

Solution for 3(2x^2-5x-33)=0 equation:

3(2x^2-5x-33)=0

We multiply parentheses

6x^2-15x-99=0

a = 6; b = -15; c = -99;
Δ = b2-4ac
Δ = -152-4·6·(-99)
Δ = 2601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2601}=51$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-51}{2*6}=\frac{-36}{12}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+51}{2*6}=\frac{66}{12}
=5+1/2
$