9=b(9/b)

Solution for 9=b(9/b) equation:

9=b(9/b)

We move all terms to the left:

9-(b(9/b))=0


Domain of the equation: b))!=0

b!=0/1

b!=0

b∈R


We add all the numbers together, and all the variables

-(b(+9/b))+9=0

We multiply all the terms by the denominator


-(b(+9+9*b))=0


We calculate terms in parentheses: -(b(+9+9*b)), so:

b(+9+9*b)

We add all the numbers together, and all the variables

b(9b+9)

We multiply parentheses

9b^2+9b

Back to the equation:

-(9b^2+9b)


We get rid of parentheses

-9b^2-9b=0

a = -9; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·(-9)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*-9}=\frac{0}{-18}
=0
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*-9}=\frac{18}{-18}
=-1
$