4/3x-13/3=(4x-13)/3

Solution for 4/3x-13/3=(4x-13)/3 equation:

4/3x-13/3=(4x-13)/3

We move all terms to the left:

4/3x-13/3-((4x-13)/3)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We calculate fractions


()/27x^2+(-((4x-13)*3x)/27x^2=0

We multiply all the terms by the denominator


(-((4x-13)*3x)+()=0


We calculate terms in parentheses: +(-((4x-13)*3x)+(), so:

-((4x-13)*3x)+(

We add all the numbers together, and all the variables

-((4x-13)*3x)


We calculate terms in parentheses: -((4x-13)*3x), so:

(4x-13)*3x

We multiply parentheses

12x^2-39x

Back to the equation:

-(12x^2-39x)


We get rid of parentheses

-12x^2+39x

Back to the equation:

+(-12x^2+39x)


We get rid of parentheses

-12x^2+39x=0

a = -12; b = 39; c = 0;
Δ = b2-4ac
Δ = 392-4·(-12)·0
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-39}{2*-12}=\frac{-78}{-24}
=3+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+39}{2*-12}=\frac{0}{-24}
=0
$