3b+b^2=1

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

3b+b^2=1

We move all terms to the left:

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

a = 1; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·1·(-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{-(3)-\sqrt{13}}{2*1}=\frac{-3-\sqrt{13}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{13}}{2*1}=\frac{-3+\sqrt{13}}{2}
$