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

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

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

We move all terms to the left:

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

We multiply parentheses

12b^2+6b+8b-(2(3b+2))+3+1=0


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

2(3b+2)

We multiply parentheses

6b+4

Back to the equation:

-(6b+4)


We add all the numbers together, and all the variables

12b^2+14b-(6b+4)+4=0

We get rid of parentheses

12b^2+14b-6b-4+4=0

We add all the numbers together, and all the variables

12b^2+8b=0

a = 12; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·12·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*12}=\frac{-16}{24}
=-2/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*12}=\frac{0}{24}
=0
$