b-b(b=2)

Solution for b-b(b=2) equation:

b-b(b=2)

We move all terms to the left:

b-b(b-(2))=0

We multiply parentheses

-b^2+b+2b=0

We add all the numbers together, and all the variables

-1b^2+3b=0

a = -1; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-1)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-1}=\frac{-6}{-2}
=+3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-1}=\frac{0}{-2}
=0
$