b^2=5+b

Solution for b^2=5+b equation:

b^2=5+b

We move all terms to the left:

b^2-(5+b)=0

We add all the numbers together, and all the variables

b^2-(b+5)=0

We get rid of parentheses

b^2-b-5=0

We add all the numbers together, and all the variables

b^2-1b-5=0

a = 1; b = -1; c = -5;
Δ = b2-4ac
Δ = -12-4·1·(-5)
Δ = 21
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{-(-1)-\sqrt{21}}{2*1}=\frac{1-\sqrt{21}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{21}}{2*1}=\frac{1+\sqrt{21}}{2}
$