-(3x)^2+6=-3

Solution for -(3x)^2+6=-3 equation:

-(3x)^2+6=-3

We move all terms to the left:

-(3x)^2+6-(-3)=0

We add all the numbers together, and all the variables

-3x^2+9=0

a = -3; b = 0; c = +9;
Δ = b2-4ac
Δ = 02-4·(-3)·9
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{-(0)-6\sqrt{3}}{2*-3}=\frac{0-6\sqrt{3}}{-6}
=-\frac{6\sqrt{3}}{-6}
=-\frac{\sqrt{3}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{3}}{2*-3}=\frac{0+6\sqrt{3}}{-6}
=\frac{6\sqrt{3}}{-6}
=\frac{\sqrt{3}}{-1}
$