(1/3)(3x+6)=2x+5

Solution for (1/3)(3x+6)=2x+5 equation:

(1/3)(3x+6)=2x+5

We move all terms to the left:

(1/3)(3x+6)-(2x+5)=0


Domain of the equation: 3)(3x+6)!=0

x∈R


We add all the numbers together, and all the variables

(+1/3)(3x+6)-(2x+5)=0

We get rid of parentheses

(+1/3)(3x+6)-2x-5=0

We multiply parentheses ..

(+3x^2+1/3*6)-2x-5=0

We multiply all the terms by the denominator


(+3x^2+1-2x*3*6)-5*3*6)=0

We add all the numbers together, and all the variables

(+3x^2+1-2x*3*6)=0

We get rid of parentheses

3x^2-2x*3*6+1=0

Wy multiply elements

3x^2-36x*6+1=0

Wy multiply elements

3x^2-216x+1=0

a = 3; b = -216; c = +1;
Δ = b2-4ac
Δ = -2162-4·3·1
Δ = 46644
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:

$\sqrt{\Delta}=\sqrt{46644}=\sqrt{676*69}=\sqrt{676}*\sqrt{69}=26\sqrt{69}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-216)-26\sqrt{69}}{2*3}=\frac{216-26\sqrt{69}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-216)+26\sqrt{69}}{2*3}=\frac{216+26\sqrt{69}}{6}
$