3b(b+1)=4b+1

Solution for 3b(b+1)=4b+1 equation:

3b(b+1)=4b+1

We move all terms to the left:

3b(b+1)-(4b+1)=0

We multiply parentheses

3b^2+3b-(4b+1)=0

We get rid of parentheses

3b^2+3b-4b-1=0

We add all the numbers together, and all the variables

3b^2-1b-1=0

a = 3; b = -1; c = -1;
Δ = b2-4ac
Δ = -12-4·3·(-1)
Δ = 13
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{13}}{2*3}=\frac{1-\sqrt{13}}{6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{13}}{2*3}=\frac{1+\sqrt{13}}{6}
$