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

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

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

a = -3; b = -3; c = +36;
Δ = b2-4ac
Δ = -32-4·(-3)·36
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-21}{2*-3}=\frac{-18}{-6}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+21}{2*-3}=\frac{24}{-6}
=-4
$