3b=3b^2-3b

Solution for 3b=3b^2-3b equation:

3b=3b^2-3b

We move all terms to the left:

3b-(3b^2-3b)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3b^2+6b=0

a = -3; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-3)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-3}=\frac{-12}{-6}
=+2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-3}=\frac{0}{-6}
=0
$