28-4b=(b+1)*(b+1)

Solution for 28-4b=(b+1)*(b+1) equation:

28-4b=(b+1)(b+1)

We move all terms to the left:

28-4b-((b+1)(b+1))=0

We multiply parentheses ..

-((+b^2+b+b+1))-4b+28=0


We calculate terms in parentheses: -((+b^2+b+b+1)), so:

(+b^2+b+b+1)

We get rid of parentheses

b^2+b+b+1

We add all the numbers together, and all the variables

b^2+2b+1

Back to the equation:

-(b^2+2b+1)


We add all the numbers together, and all the variables

-4b-(b^2+2b+1)+28=0

We get rid of parentheses

-b^2-4b-2b-1+28=0

We add all the numbers together, and all the variables

-1b^2-6b+27=0

a = -1; b = -6; c = +27;
Δ = b2-4ac
Δ = -62-4·(-1)·27
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-12}{2*-1}=\frac{-6}{-2}
=+3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+12}{2*-1}=\frac{18}{-2}
=-9
$