W^2+5w=w^2

Solution for W^2+5w=w^2 equation:

^2+5W=W^2

We move all terms to the left:

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

determiningTheFunctionDomain
-W^2+5W+^2=0

We add all the numbers together, and all the variables

-1W^2+5W=0

a = -1; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-1)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$W_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$W_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-1}=\frac{-10}{-2}
=+5
$
$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-1}=\frac{0}{-2}
=0
$