1=3b-b^2

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

1=3b-b^2

We move all terms to the left:

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

We get rid of parentheses

b^2-3b+1=0

a = 1; b = -3; c = +1;
Δ = b2-4ac
Δ = -32-4·1·1
Δ = 5
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{-(-3)-\sqrt{5}}{2*1}=\frac{3-\sqrt{5}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{5}}{2*1}=\frac{3+\sqrt{5}}{2}
$