Y=3(x-2)(x+4)

Solution for Y=3(x-2)(x+4) equation:

=3(Y-2)(Y+4)

We move all terms to the left:

-(3(Y-2)(Y+4))=0

We multiply parentheses ..

-(3(+Y^2+4Y-2Y-8))=0


We calculate terms in parentheses: -(3(+Y^2+4Y-2Y-8)), so:

3(+Y^2+4Y-2Y-8)

We multiply parentheses

3Y^2+12Y-6Y-24

We add all the numbers together, and all the variables

3Y^2+6Y-24

Back to the equation:

-(3Y^2+6Y-24)


We get rid of parentheses

-3Y^2-6Y+24=0

a = -3; b = -6; c = +24;
Δ = b2-4ac
Δ = -62-4·(-3)·24
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-18}{2*-3}=\frac{-12}{-6}
=+2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+18}{2*-3}=\frac{24}{-6}
=-4
$