Y=(x-4)(x+8)

Solution for Y=(x-4)(x+8) equation:

=(Y-4)(Y+8)

We move all terms to the left:

-((Y-4)(Y+8))=0

We multiply parentheses ..

-((+Y^2+8Y-4Y-32))=0


We calculate terms in parentheses: -((+Y^2+8Y-4Y-32)), so:

(+Y^2+8Y-4Y-32)

We get rid of parentheses

Y^2+8Y-4Y-32

We add all the numbers together, and all the variables

Y^2+4Y-32

Back to the equation:

-(Y^2+4Y-32)


We get rid of parentheses

-Y^2-4Y+32=0

We add all the numbers together, and all the variables

-1Y^2-4Y+32=0

a = -1; b = -4; c = +32;
Δ = b2-4ac
Δ = -42-4·(-1)·32
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{-(-4)-12}{2*-1}=\frac{-8}{-2}
=+4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*-1}=\frac{16}{-2}
=-8
$