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

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

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

a = -3; b = 18; c = +9;
Δ = b2-4ac
Δ = 182-4·(-3)·9
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*-3}=\frac{-18-12\sqrt{3}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+12\sqrt{3}}{2*-3}=\frac{-18+12\sqrt{3}}{-6}
$