w(w-5)-w-22+(w+5)(w-5)=w

Solution for w(w-5)-w-22+(w+5)(w-5)=w equation:

w(w-5)-w-22+(w+5)(w-5)=w

We move all terms to the left:

w(w-5)-w-22+(w+5)(w-5)-(w)=0

We add all the numbers together, and all the variables

-2w+w(w-5)+(w+5)(w-5)-22=0

We use the square of the difference formula

w^2-2w+w(w-5)-25-22=0

We multiply parentheses

w^2+w^2-2w-5w-25-22=0

We add all the numbers together, and all the variables

2w^2-7w-47=0

a = 2; b = -7; c = -47;
Δ = b2-4ac
Δ = -72-4·2·(-47)
Δ = 425
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:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-5\sqrt{17}}{2*2}=\frac{7-5\sqrt{17}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+5\sqrt{17}}{2*2}=\frac{7+5\sqrt{17}}{4}
$