-3x^2+18x+6=0

Solution for -3x^2+18x+6=0 equation:

-3x^2+18x+6=0

a = -3; b = 18; c = +6;
Δ = b2-4ac
Δ = 182-4·(-3)·6
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{11}}{2*-3}=\frac{-18-6\sqrt{11}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{11}}{2*-3}=\frac{-18+6\sqrt{11}}{-6}
$