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

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

=(Y+2)(Y-4)

We move all terms to the left:

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

We multiply parentheses ..

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


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

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

We get rid of parentheses

Y^2-4Y+2Y-8

We add all the numbers together, and all the variables

Y^2-2Y-8

Back to the equation:

-(Y^2-2Y-8)


We get rid of parentheses

-Y^2+2Y+8=0

We add all the numbers together, and all the variables

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