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

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

=(4-Y)(Y+2)

We move all terms to the left:

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

We add all the numbers together, and all the variables

-((-1Y+4)(Y+2))=0

We multiply parentheses ..

-((-1Y^2-2Y+4Y+8))=0


We calculate terms in parentheses: -((-1Y^2-2Y+4Y+8)), so:

(-1Y^2-2Y+4Y+8)

We get rid of parentheses

-1Y^2-2Y+4Y+8

We add all the numbers together, and all the variables

-1Y^2+2Y+8

Back to the equation:

-(-1Y^2+2Y+8)


We get rid of parentheses

1Y^2-2Y-8=0

We add all the numbers together, and all the variables

Y^2-2Y-8=0

a = 1; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·1·(-8)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*1}=\frac{-4}{2}
=-2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*1}=\frac{8}{2}
=4
$