-3x2-6x+12=0

Solution for -3x2-6x+12=0 equation:

-3x^2-6x+12=0

a = -3; b = -6; c = +12;
Δ = b2-4ac
Δ = -62-4·(-3)·12
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6\sqrt{5}}{2*-3}=\frac{6-6\sqrt{5}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6\sqrt{5}}{2*-3}=\frac{6+6\sqrt{5}}{-6}
$