-3x^2-18x+36=0

Solution for -3x^2-18x+36=0 equation:

-3x^2-18x+36=0

a = -3; b = -18; c = +36;
Δ = b2-4ac
Δ = -182-4·(-3)·36
Δ = 756
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:

$\sqrt{\Delta}=\sqrt{756}=\sqrt{36*21}=\sqrt{36}*\sqrt{21}=6\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{21}}{2*-3}=\frac{18-6\sqrt{21}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{21}}{2*-3}=\frac{18+6\sqrt{21}}{-6}
$