33=3x(2x+1)

Solution for 33=3x(2x+1) equation:

33=3x(2x+1)

We move all terms to the left:

33-(3x(2x+1))=0


We calculate terms in parentheses: -(3x(2x+1)), so:

3x(2x+1)

We multiply parentheses

6x^2+3x

Back to the equation:

-(6x^2+3x)


We get rid of parentheses

-6x^2-3x+33=0

a = -6; b = -3; c = +33;
Δ = b2-4ac
Δ = -32-4·(-6)·33
Δ = 801
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:

$\sqrt{\Delta}=\sqrt{801}=\sqrt{9*89}=\sqrt{9}*\sqrt{89}=3\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{89}}{2*-6}=\frac{3-3\sqrt{89}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{89}}{2*-6}=\frac{3+3\sqrt{89}}{-12}
$