33=3x^2-4x

Solution for 33=3x^2-4x equation:

33=3x^2-4x

We move all terms to the left:

33-(3x^2-4x)=0

We get rid of parentheses

-3x^2+4x+33=0

a = -3; b = 4; c = +33;
Δ = b2-4ac
Δ = 42-4·(-3)·33
Δ = 412
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:

$\sqrt{\Delta}=\sqrt{412}=\sqrt{4*103}=\sqrt{4}*\sqrt{103}=2\sqrt{103}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{103}}{2*-3}=\frac{-4-2\sqrt{103}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{103}}{2*-3}=\frac{-4+2\sqrt{103}}{-6}
$