0=9x^2+18x-3

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

0=9x^2+18x-3

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

a = -9; b = -18; c = +3;
Δ = b2-4ac
Δ = -182-4·(-9)·3
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-12\sqrt{3}}{2*-9}=\frac{18-12\sqrt{3}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+12\sqrt{3}}{2*-9}=\frac{18+12\sqrt{3}}{-18}
$