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

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

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

a = -3; b = -3; c = +6;
Δ = b2-4ac
Δ = -32-4·(-3)·6
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9}{2*-3}=\frac{-6}{-6}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9}{2*-3}=\frac{12}{-6}
=-2
$