(2/3)(6x+3)=(1/4x)+.5

Solution for (2/3)(6x+3)=(1/4x)+.5 equation:

(2/3)(6x+3)=(1/4x)+.5

We move all terms to the left:

(2/3)(6x+3)-((1/4x)+.5)=0


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

x∈R



Domain of the equation: 4x)+.5)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+2/3)(6x+3)-((+1/4x)+.5)=0

We multiply parentheses ..

(+12x^2+2/3*3)-((+1/4x)+.5)=0

We calculate fractions


(12x^2+8x)/36x^2+()/36x^2=0

We multiply all the terms by the denominator


(12x^2+8x)+()=0

We add all the numbers together, and all the variables

(12x^2+8x)=0

We get rid of parentheses

12x^2+8x=0

a = 12; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·12·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*12}=\frac{-16}{24}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*12}=\frac{0}{24}
=0
$