-1-3b(b+3)=5(2-3b)

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Solution for -1-3b(b+3)=5(2-3b) equation: 



-1-3b(b+3)=5(2-3b)

We move all terms to the left:

-1-3b(b+3)-(5(2-3b))=0

We add all the numbers together, and all the variables

-3b(b+3)-(5(-3b+2))-1=0

We multiply parentheses

-3b^2-9b-(5(-3b+2))-1=0



We calculate terms in parentheses: -(5(-3b+2)), so:

5(-3b+2)

We multiply parentheses

-15b+10

Back to the equation:

-(-15b+10)



We get rid of parentheses

-3b^2-9b+15b-10-1=0

We add all the numbers together, and all the variables

-3b^2+6b-11=0

a = -3; b = 6; c = -11;
Δ = b2-4ac
Δ = 62-4·(-3)·(-11)
Δ = -96
Delta is less than zero, so there is no solution for the equation

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