2x+1=6(1/3x+3)-8

Solution for 2x+1=6(1/3x+3)-8 equation:

2x+1=6(1/3x+3)-8

We move all terms to the left:

2x+1-(6(1/3x+3)-8)=0


Domain of the equation: 3x+3)-8)!=0

x∈R


We multiply all the terms by the denominator


2x*3x+1*3x+3)-8)-(6(1+3)-8)=0

We add all the numbers together, and all the variables

2x*3x+1*3x+3)-8)-(64-8)=0

We add all the numbers together, and all the variables

2x*3x+1*3x=0

Wy multiply elements

6x^2+3x=0

a = 6; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·6·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*6}=\frac{-6}{12}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*6}=\frac{0}{12}
=0
$