3b^2=-2b+1

Solution for 3b^2=-2b+1 equation:

3b^2=-2b+1

We move all terms to the left:

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

We get rid of parentheses

3b^2+2b-1=0

a = 3; b = 2; c = -1;
Δ = b2-4ac
Δ = 22-4·3·(-1)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4}{2*3}=\frac{-6}{6}
=-1
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*3}=\frac{2}{6}
=1/3
$