b^2=21+4b

Solution for b^2=21+4b equation:

b^2=21+4b

We move all terms to the left:

b^2-(21+4b)=0

We add all the numbers together, and all the variables

b^2-(4b+21)=0

We get rid of parentheses

b^2-4b-21=0

a = 1; b = -4; c = -21;
Δ = b2-4ac
Δ = -42-4·1·(-21)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-10}{2*1}=\frac{-6}{2}
=-3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+10}{2*1}=\frac{14}{2}
=7
$