x=5x-9+((2/3)*(5x-9))+x

Solution for x=5x-9+((2/3)*(5x-9))+x equation:

x=5x-9+((2/3)(5x-9))+x

We move all terms to the left:

x-(5x-9+((2/3)(5x-9))+x)=0


Domain of the equation: 3)(5x-9))+x)!=0

x∈R


We add all the numbers together, and all the variables

x-(5x-9+((+2/3)(5x-9))+x)=0

We multiply parentheses ..

-(5x-9+((+10x^2+2/3*-9))+x)+x=0

We multiply all the terms by the denominator


-(5x-9+((+10x^2+2+x*3*-9))+x)=0


We calculate terms in parentheses: -(5x-9+((+10x^2+2+x*3*-9))+x), so:

5x-9+((+10x^2+2+x*3*-9))+x

determiningTheFunctionDomain
((+10x^2+2+x*3*-9))+5x+x-9

We add all the numbers together, and all the variables

((+10x^2+2+x*3*-9))+6x-9


We calculate terms in parentheses: +((+10x^2+2+x*3*-9)), so:

(+10x^2+2+x*3*-9)

We get rid of parentheses

10x^2+x*3*+2-9

We add all the numbers together, and all the variables

10x^2+x*3*-7

Wy multiply elements

10x^2+3x^2-7

We add all the numbers together, and all the variables

13x^2-7

Back to the equation:

+(13x^2-7)


We add all the numbers together, and all the variables

6x+(13x^2-7)-9

We get rid of parentheses

13x^2+6x-7-9

We add all the numbers together, and all the variables

13x^2+6x-16

Back to the equation:

-(13x^2+6x-16)


We get rid of parentheses

-13x^2-6x+16=0

a = -13; b = -6; c = +16;
Δ = b2-4ac
Δ = -62-4·(-13)·16
Δ = 868
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{868}=\sqrt{4*217}=\sqrt{4}*\sqrt{217}=2\sqrt{217}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{217}}{2*-13}=\frac{6-2\sqrt{217}}{-26}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{217}}{2*-13}=\frac{6+2\sqrt{217}}{-26}
$