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

Solution for 0=3x^2+3x-33/4 equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

-12x^2-12x+33=0

a = -12; b = -12; c = +33;
Δ = b2-4ac
Δ = -122-4·(-12)·33
Δ = 1728
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:

$\sqrt{\Delta}=\sqrt{1728}=\sqrt{576*3}=\sqrt{576}*\sqrt{3}=24\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-24\sqrt{3}}{2*-12}=\frac{12-24\sqrt{3}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+24\sqrt{3}}{2*-12}=\frac{12+24\sqrt{3}}{-24}
$