0=(x-4)(x-4)

Solution for 0=(x-4)(x-4) equation:

0=(x-4)(x-4)

We move all terms to the left:

0-((x-4)(x-4))=0

We add all the numbers together, and all the variables

-((x-4)(x-4))=0

We multiply parentheses ..

-((+x^2-4x-4x+16))=0


We calculate terms in parentheses: -((+x^2-4x-4x+16)), so:

(+x^2-4x-4x+16)

We get rid of parentheses

x^2-4x-4x+16

We add all the numbers together, and all the variables

x^2-8x+16

Back to the equation:

-(x^2-8x+16)


We get rid of parentheses

-x^2+8x-16=0

We add all the numbers together, and all the variables

-1x^2+8x-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:
$x=\frac{-b}{2a}=\frac{-8}{-2}=+4$