48=(w)(w+2)(2)

Solution for 48=(w)(w+2)(2) equation:

48=(w)(w+2)(2)

We move all terms to the left:

48-((w)(w+2)(2))=0


We calculate terms in parentheses: -(w(w+2)2), so:

w(w+2)2

We multiply parentheses

2w^2+4w

Back to the equation:

-(2w^2+4w)


We get rid of parentheses

-2w^2-4w+48=0

a = -2; b = -4; c = +48;
Δ = b2-4ac
Δ = -42-4·(-2)·48
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-20}{2*-2}=\frac{-16}{-4}
=+4
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+20}{2*-2}=\frac{24}{-4}
=-6
$