b=b^2+b-56

Solution for b=b^2+b-56 equation:

b=b^2+b-56

We move all terms to the left:

b-(b^2+b-56)=0

We get rid of parentheses

-b^2+b-b+56=0

We add all the numbers together, and all the variables

-1b^2+56=0

a = -1; b = 0; c = +56;
Δ = b2-4ac
Δ = 02-4·(-1)·56
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{14}}{2*-1}=\frac{0-4\sqrt{14}}{-2}
=-\frac{4\sqrt{14}}{-2}
=-\frac{2\sqrt{14}}{-1}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{14}}{2*-1}=\frac{0+4\sqrt{14}}{-2}
=\frac{4\sqrt{14}}{-2}
=\frac{2\sqrt{14}}{-1}
$