b^2=5b+104

Solution for b^2=5b+104 equation:

b^2=5b+104

We move all terms to the left:

b^2-(5b+104)=0

We get rid of parentheses

b^2-5b-104=0

a = 1; b = -5; c = -104;
Δ = b2-4ac
Δ = -52-4·1·(-104)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-21}{2*1}=\frac{-16}{2}
=-8
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+21}{2*1}=\frac{26}{2}
=13
$