(3n+3)(n-2)=0

Solution for (3n+3)(n-2)=0 equation:

(3n+3)(n-2)=0

We multiply parentheses ..

(+3n^2-6n+3n-6)=0

We get rid of parentheses

3n^2-6n+3n-6=0

We add all the numbers together, and all the variables

3n^2-3n-6=0

a = 3; b = -3; c = -6;
Δ = b2-4ac
Δ = -32-4·3·(-6)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9}{2*3}=\frac{-6}{6}
=-1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9}{2*3}=\frac{12}{6}
=2
$