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

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

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

a = -3; b = 18; c = -18;
Δ = b2-4ac
Δ = 182-4·(-3)·(-18)
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{3}}{2*-3}=\frac{-18-6\sqrt{3}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{3}}{2*-3}=\frac{-18+6\sqrt{3}}{-6}
$