-3c(c-4)-2c-8=9(c+2)+1

Solution for -3c(c-4)-2c-8=9(c+2)+1 equation:

-3c(c-4)-2c-8=9(c+2)+1

We move all terms to the left:

-3c(c-4)-2c-8-(9(c+2)+1)=0

We add all the numbers together, and all the variables

-2c-3c(c-4)-(9(c+2)+1)-8=0

We multiply parentheses

-3c^2-2c+12c-(9(c+2)+1)-8=0


We calculate terms in parentheses: -(9(c+2)+1), so:

9(c+2)+1

We multiply parentheses

9c+18+1

We add all the numbers together, and all the variables

9c+19

Back to the equation:

-(9c+19)


We add all the numbers together, and all the variables

-3c^2+10c-(9c+19)-8=0

We get rid of parentheses

-3c^2+10c-9c-19-8=0

We add all the numbers together, and all the variables

-3c^2+c-27=0

a = -3; b = 1; c = -27;
Δ = b2-4ac
Δ = 12-4·(-3)·(-27)
Δ = -323
Delta is less than zero, so there is no solution for the equation