-3x^2=39x+120

Solution for -3x^2=39x+120 equation:

-3x^2=39x+120

We move all terms to the left:

-3x^2-(39x+120)=0

We get rid of parentheses

-3x^2-39x-120=0

a = -3; b = -39; c = -120;
Δ = b2-4ac
Δ = -392-4·(-3)·(-120)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-9}{2*-3}=\frac{30}{-6}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+9}{2*-3}=\frac{48}{-6}
=-8
$