-3y^2-120y+1200=0

Solution for -3y^2-120y+1200=0 equation:

-3y^2-120y+1200=0

a = -3; b = -120; c = +1200;
Δ = b2-4ac
Δ = -1202-4·(-3)·1200
Δ = 28800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{28800}=\sqrt{14400*2}=\sqrt{14400}*\sqrt{2}=120\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-120\sqrt{2}}{2*-3}=\frac{120-120\sqrt{2}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+120\sqrt{2}}{2*-3}=\frac{120+120\sqrt{2}}{-6}
$