-3x^2-36x+9=0

Solution for -3x^2-36x+9=0 equation:

-3x^2-36x+9=0

a = -3; b = -36; c = +9;
Δ = b2-4ac
Δ = -362-4·(-3)·9
Δ = 1404
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:

$\sqrt{\Delta}=\sqrt{1404}=\sqrt{36*39}=\sqrt{36}*\sqrt{39}=6\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-6\sqrt{39}}{2*-3}=\frac{36-6\sqrt{39}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+6\sqrt{39}}{2*-3}=\frac{36+6\sqrt{39}}{-6}
$