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

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

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

a = -3; b = 9; c = +18;
Δ = b2-4ac
Δ = 92-4·(-3)·18
Δ = 297
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:

$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{33}}{2*-3}=\frac{-9-3\sqrt{33}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{33}}{2*-3}=\frac{-9+3\sqrt{33}}{-6}
$