2+3b-b^2=3

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

2+3b-b^2=3

We move all terms to the left:

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

We add all the numbers together, and all the variables

-1b^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}
$