(2^2-x/2)=(1/128)

Solution for (2^2-x/2)=(1/128) equation:

(2^2-x/2)=(1/128)

We move all terms to the left:

(2^2-x/2)-((1/128))=0

We add all the numbers together, and all the variables

(2^2-x/2)-((+1/128))=0

We get rid of parentheses

-x/2+2^2-((+1/128))=0

We calculate fractions


(-128x^2)/()+2^2+()/()=0

We add all the numbers together, and all the variables

(-128x^2)/()+5=0

We multiply all the terms by the denominator


(-128x^2)+5*()=0

We add all the numbers together, and all the variables

(-128x^2)=0

We get rid of parentheses

-128x^2=0

a = -128; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-128)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{-256}=0$