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

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

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

We move all terms to the left:

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


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

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply parentheses ..

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

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

We get rid of parentheses

2x^2-x*3*9+2=0

Wy multiply elements

2x^2-27x*9+2=0

Wy multiply elements

2x^2-243x+2=0

a = 2; b = -243; c = +2;
Δ = b2-4ac
Δ = -2432-4·2·2
Δ = 59033
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-243)-\sqrt{59033}}{2*2}=\frac{243-\sqrt{59033}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-243)+\sqrt{59033}}{2*2}=\frac{243+\sqrt{59033}}{4}
$