((8x^2)/(9-x))=9+x

Solution for ((8x^2)/(9-x))=9+x equation:

((8x^2)/(9-x))=9+x

We move all terms to the left:

((8x^2)/(9-x))-(9+x)=0


Domain of the equation: (9-x))!=0

We move all terms containing x to the left, all other terms to the right

-x)!=-9

x!=-9/1

x!=-9

x∈R


We add all the numbers together, and all the variables

(8x^2/(-1x+9))-(x+9)=0

We get rid of parentheses

(8x^2/(-1x+9))-x-9=0

We multiply all the terms by the denominator


(8x^2-x*(-1x+9))-9*(-1x+9))=0


We calculate terms in parentheses: +(8x^2-x*(-1x+9)), so:

8x^2-x*(-1x+9)

We multiply parentheses

8x^2+1x^2-9x

We add all the numbers together, and all the variables

9x^2-9x

Back to the equation:

+(9x^2-9x)


We multiply parentheses

(9x^2-9x)+9x-=0

We get rid of parentheses

9x^2-9x+9x-=0

We add all the numbers together, and all the variables

9x^2=0

a = 9; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·9·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$x=\frac{-b}{2a}=\frac{0}{18}=0$