36=b^2+5b

Solution for 36=b^2+5b equation:

36=b^2+5b

We move all terms to the left:

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

We get rid of parentheses

-b^2-5b+36=0

We add all the numbers together, and all the variables

-1b^2-5b+36=0

a = -1; b = -5; c = +36;
Δ = b2-4ac
Δ = -52-4·(-1)·36
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*-1}=\frac{-8}{-2}
=+4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*-1}=\frac{18}{-2}
=-9
$