120=(3x^2)+(3x)-6

Solution for 120=(3x^2)+(3x)-6 equation:

120=(3x^2)+(3x)-6

We move all terms to the left:

120-((3x^2)+(3x)-6)=0

We get rid of parentheses

-3x^2-3x+6+120=0

We add all the numbers together, and all the variables

-3x^2-3x+126=0

a = -3; b = -3; c = +126;
Δ = b2-4ac
Δ = -32-4·(-3)·126
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-39}{2*-3}=\frac{-36}{-6}
=+6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+39}{2*-3}=\frac{42}{-6}
=-7
$