320=(w+8)(w+4)

Solution for 320=(w+8)(w+4) equation:

320=(w+8)(w+4)

We move all terms to the left:

320-((w+8)(w+4))=0

We multiply parentheses ..

-((+w^2+4w+8w+32))+320=0


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

(+w^2+4w+8w+32)

We get rid of parentheses

w^2+4w+8w+32

We add all the numbers together, and all the variables

w^2+12w+32

Back to the equation:

-(w^2+12w+32)


We get rid of parentheses

-w^2-12w-32+320=0

We add all the numbers together, and all the variables

-1w^2-12w+288=0

a = -1; b = -12; c = +288;
Δ = b2-4ac
Δ = -122-4·(-1)·288
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-36}{2*-1}=\frac{-24}{-2}
=+12
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+36}{2*-1}=\frac{48}{-2}
=-24
$