3c(c-1)+2(3c+1)=-(3c+1)

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Solution for 3c(c-1)+2(3c+1)=-(3c+1) equation: 



3c(c-1)+2(3c+1)=-(3c+1)

We move all terms to the left:

3c(c-1)+2(3c+1)-(-(3c+1))=0

We multiply parentheses

3c^2-3c+6c-(-(3c+1))+2=0



We calculate terms in parentheses: -(-(3c+1)), so:

-(3c+1)

We get rid of parentheses

-3c-1

Back to the equation:

-(-3c-1)



We add all the numbers together, and all the variables

3c^2+3c-(-3c-1)+2=0

We get rid of parentheses

3c^2+3c+3c+1+2=0

We add all the numbers together, and all the variables

3c^2+6c+3=0

a = 3; b = 6; c = +3;
Δ = b2-4ac
Δ = 62-4·3·3
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$c=\frac{-b}{2a}=\frac{-6}{6}=-1$

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