W(w+5)+28=(w+2)(w+5)

Solution for W(w+5)+28=(w+2)(w+5) equation:

(W+5)+28=(W+2)(W+5)

We move all terms to the left:

(W+5)+28-((W+2)(W+5))=0

We get rid of parentheses

W-((W+2)(W+5))+5+28=0

We multiply parentheses ..

-((+W^2+5W+2W+10))+W+5+28=0


We calculate terms in parentheses: -((+W^2+5W+2W+10)), so:

(+W^2+5W+2W+10)

We get rid of parentheses

W^2+5W+2W+10

We add all the numbers together, and all the variables

W^2+7W+10

Back to the equation:

-(W^2+7W+10)


We add all the numbers together, and all the variables

W-(W^2+7W+10)+33=0

We get rid of parentheses

-W^2+W-7W-10+33=0

We add all the numbers together, and all the variables

-1W^2-6W+23=0

a = -1; b = -6; c = +23;
Δ = b2-4ac
Δ = -62-4·(-1)·23
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$W_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$W_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-8\sqrt{2}}{2*-1}=\frac{6-8\sqrt{2}}{-2}
$
$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+8\sqrt{2}}{2*-1}=\frac{6+8\sqrt{2}}{-2}
$