5/2(2c+18)=1/5(25c+30)

Solution for 5/2(2c+18)=1/5(25c+30) equation:

5/2(2c+18)=1/5(25c+30)

We move all terms to the left:

5/2(2c+18)-(1/5(25c+30))=0


Domain of the equation: 2(2c+18)!=0

c∈R



Domain of the equation: 5(25c+30))!=0

c∈R


We calculate fractions


(25c2/(2(2c+18)*5(25c+30)))+(-2c2/(2(2c+18)*5(25c+30)))=0


We calculate terms in parentheses: +(25c2/(2(2c+18)*5(25c+30))), so:

25c2/(2(2c+18)*5(25c+30))

We multiply all the terms by the denominator


25c2

We add all the numbers together, and all the variables

25c^2

Back to the equation:

+(25c^2)



We calculate terms in parentheses: +(-2c2/(2(2c+18)*5(25c+30))), so:

-2c2/(2(2c+18)*5(25c+30))

We multiply all the terms by the denominator


-2c2

We add all the numbers together, and all the variables

-2c^2

Back to the equation:

+(-2c^2)


We add all the numbers together, and all the variables

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

We get rid of parentheses

25c^2-2c^2=0

We add all the numbers together, and all the variables

23c^2=0

a = 23; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·23·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$c=\frac{-b}{2a}=\frac{0}{46}=0$