Y=(3x-3)(x+3)

Solution for Y=(3x-3)(x+3) equation:

=(3Y-3)(Y+3)

We move all terms to the left:

-((3Y-3)(Y+3))=0

We multiply parentheses ..

-((+3Y^2+9Y-3Y-9))=0


We calculate terms in parentheses: -((+3Y^2+9Y-3Y-9)), so:

(+3Y^2+9Y-3Y-9)

We get rid of parentheses

3Y^2+9Y-3Y-9

We add all the numbers together, and all the variables

3Y^2+6Y-9

Back to the equation:

-(3Y^2+6Y-9)


We get rid of parentheses

-3Y^2-6Y+9=0

a = -3; b = -6; c = +9;
Δ = b2-4ac
Δ = -62-4·(-3)·9
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-12}{2*-3}=\frac{-6}{-6}
=1
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+12}{2*-3}=\frac{18}{-6}
=-3
$