28=w(w+3)

Solution for 28=w(w+3) equation:

28=w(w+3)

We move all terms to the left:

28-(w(w+3))=0


We calculate terms in parentheses: -(w(w+3)), so:

w(w+3)

We multiply parentheses

w^2+3w

Back to the equation:

-(w^2+3w)


We get rid of parentheses

-w^2-3w+28=0

We add all the numbers together, and all the variables

-1w^2-3w+28=0

a = -1; b = -3; c = +28;
Δ = b2-4ac
Δ = -32-4·(-1)·28
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-11}{2*-1}=\frac{-8}{-2}
=+4
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+11}{2*-1}=\frac{14}{-2}
=-7
$