0=3x^2-33x+90

Solution for 0=3x^2-33x+90 equation:

0=3x^2-33x+90

We move all terms to the left:

0-(3x^2-33x+90)=0

We add all the numbers together, and all the variables

-(3x^2-33x+90)=0

We get rid of parentheses

-3x^2+33x-90=0

a = -3; b = 33; c = -90;
Δ = b2-4ac
Δ = 332-4·(-3)·(-90)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-3}{2*-3}=\frac{-36}{-6}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+3}{2*-3}=\frac{-30}{-6}
=+5
$