Y=3(x+3)(x-2)

Solution for Y=3(x+3)(x-2) equation:

=3(Y+3)(Y-2)

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -(3(+Y^2-2Y+3Y-6)), so:

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

We multiply parentheses

3Y^2-6Y+9Y-18

We add all the numbers together, and all the variables

3Y^2+3Y-18

Back to the equation:

-(3Y^2+3Y-18)


We get rid of parentheses

-3Y^2-3Y+18=0

a = -3; b = -3; c = +18;
Δ = b2-4ac
Δ = -32-4·(-3)·18
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-15}{2*-3}=\frac{-12}{-6}
=+2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+15}{2*-3}=\frac{18}{-6}
=-3
$