3w^2-(w-5)(w-3)=3w^2+1

Solution for 3w^2-(w-5)(w-3)=3w^2+1 equation:

3w^2-(w-5)(w-3)=3w^2+1

We move all terms to the left:

3w^2-(w-5)(w-3)-(3w^2+1)=0

We get rid of parentheses

3w^2-3w^2-(w-5)(w-3)-1=0

We multiply parentheses ..

3w^2-3w^2-(+w^2-3w-5w+15)-1=0

We add all the numbers together, and all the variables

-(+w^2-3w-5w+15)-1=0

We get rid of parentheses

-w^2+3w+5w-15-1=0

We add all the numbers together, and all the variables

-1w^2+8w-16=0

a = -1; b = 8; c = -16;
Δ = b2-4ac
Δ = 82-4·(-1)·(-16)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$w=\frac{-b}{2a}=\frac{-8}{-2}=+4$