(2/3)(3x+24)=3x+18

Solution for (2/3)(3x+24)=3x+18 equation:

(2/3)(3x+24)=3x+18

We move all terms to the left:

(2/3)(3x+24)-(3x+18)=0


Domain of the equation: 3)(3x+24)!=0

x∈R


We add all the numbers together, and all the variables

(+2/3)(3x+24)-(3x+18)=0

We get rid of parentheses

(+2/3)(3x+24)-3x-18=0

We multiply parentheses ..

(+6x^2+2/3*24)-3x-18=0

We multiply all the terms by the denominator


(+6x^2+2-3x*3*24)-18*3*24)=0

We add all the numbers together, and all the variables

(+6x^2+2-3x*3*24)=0

We get rid of parentheses

6x^2-3x*3*24+2=0

Wy multiply elements

6x^2-216x*2+2=0

Wy multiply elements

6x^2-432x+2=0

a = 6; b = -432; c = +2;
Δ = b2-4ac
Δ = -4322-4·6·2
Δ = 186576
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:

$\sqrt{\Delta}=\sqrt{186576}=\sqrt{2704*69}=\sqrt{2704}*\sqrt{69}=52\sqrt{69}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-432)-52\sqrt{69}}{2*6}=\frac{432-52\sqrt{69}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-432)+52\sqrt{69}}{2*6}=\frac{432+52\sqrt{69}}{12}
$