w(4w-15)=w^2

Solution for w(4w-15)=w^2 equation:

w(4w-15)=w^2

We move all terms to the left:

w(4w-15)-(w^2)=0

determiningTheFunctionDomain
-w^2+w(4w-15)=0

We add all the numbers together, and all the variables

-1w^2+w(4w-15)=0

We multiply parentheses

-1w^2+4w^2-15w=0

We add all the numbers together, and all the variables

3w^2-15w=0

a = 3; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·3·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-15}{2*3}=\frac{0}{6}
=0
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*3}=\frac{30}{6}
=5
$