(2/3)(3x+9)=18

Solution for (2/3)(3x+9)=18 equation:

(2/3)(3x+9)=18

We move all terms to the left:

(2/3)(3x+9)-(18)=0


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

x∈R


We add all the numbers together, and all the variables

(+2/3)(3x+9)-18=0

We multiply parentheses ..

(+6x^2+2/3*9)-18=0

We multiply all the terms by the denominator


(+6x^2+2-18*3*9)=0

We get rid of parentheses

6x^2+2-18*3*9=0

We add all the numbers together, and all the variables

6x^2-484=0

a = 6; b = 0; c = -484;
Δ = b2-4ac
Δ = 02-4·6·(-484)
Δ = 11616
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:

$\sqrt{\Delta}=\sqrt{11616}=\sqrt{1936*6}=\sqrt{1936}*\sqrt{6}=44\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-44\sqrt{6}}{2*6}=\frac{0-44\sqrt{6}}{12}
=-\frac{44\sqrt{6}}{12}
=-\frac{11\sqrt{6}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+44\sqrt{6}}{2*6}=\frac{0+44\sqrt{6}}{12}
=\frac{44\sqrt{6}}{12}
=\frac{11\sqrt{6}}{3}
$