3b(2b+4)=(b^2+b-7)

Solution for 3b(2b+4)=(b^2+b-7) equation:

3b(2b+4)=(b^2+b-7)

We move all terms to the left:

3b(2b+4)-((b^2+b-7))=0

We multiply parentheses

6b^2+12b-((b^2+b-7))=0


We calculate terms in parentheses: -((b^2+b-7)), so:

(b^2+b-7)

We get rid of parentheses

b^2+b-7

Back to the equation:

-(b^2+b-7)


We get rid of parentheses

6b^2-b^2+12b-b+7=0

We add all the numbers together, and all the variables

5b^2+11b+7=0

a = 5; b = 11; c = +7;
Δ = b2-4ac
Δ = 112-4·5·7
Δ = -19
Delta is less than zero, so there is no solution for the equation