b=b^2+19b-20

Solution for b=b^2+19b-20 equation:

b=b^2+19b-20

We move all terms to the left:

b-(b^2+19b-20)=0

We get rid of parentheses

-b^2+b-19b+20=0

We add all the numbers together, and all the variables

-1b^2-18b+20=0

a = -1; b = -18; c = +20;
Δ = b2-4ac
Δ = -182-4·(-1)·20
Δ = 404
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{404}=\sqrt{4*101}=\sqrt{4}*\sqrt{101}=2\sqrt{101}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{101}}{2*-1}=\frac{18-2\sqrt{101}}{-2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{101}}{2*-1}=\frac{18+2\sqrt{101}}{-2}
$