0=(2c-5)(2c-5)

Solution for 0=(2c-5)(2c-5) equation:

0=(2c-5)(2c-5)

We move all terms to the left:

0-((2c-5)(2c-5))=0

We add all the numbers together, and all the variables

-((2c-5)(2c-5))=0

We multiply parentheses ..

-((+4c^2-10c-10c+25))=0


We calculate terms in parentheses: -((+4c^2-10c-10c+25)), so:

(+4c^2-10c-10c+25)

We get rid of parentheses

4c^2-10c-10c+25

We add all the numbers together, and all the variables

4c^2-20c+25

Back to the equation:

-(4c^2-20c+25)


We get rid of parentheses

-4c^2+20c-25=0

a = -4; b = 20; c = -25;
Δ = b2-4ac
Δ = 202-4·(-4)·(-25)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$c=\frac{-b}{2a}=\frac{-20}{-8}=2+1/2$