0=3x^2-18x+9

Solution for 0=3x^2-18x+9 equation:

0=3x^2-18x+9

We move all terms to the left:

0-(3x^2-18x+9)=0

We add all the numbers together, and all the variables

-(3x^2-18x+9)=0

We get rid of parentheses

-3x^2+18x-9=0

a = -3; b = 18; c = -9;
Δ = b2-4ac
Δ = 182-4·(-3)·(-9)
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{6}}{2*-3}=\frac{-18-6\sqrt{6}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{6}}{2*-3}=\frac{-18+6\sqrt{6}}{-6}
$