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

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

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

We move all terms to the left:

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


Domain of the equation: 5)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

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

We multiply parentheses

2x^2-((3x+1)/2)+(+4/3)=0

We get rid of parentheses

2x^2-((3x+1)/2)+4/3=0

We calculate fractions


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


We calculate terms in parentheses: +(-((3x+1)*3)/()+()/(), so:

-((3x+1)*3)/()+()/(

We add all the numbers together, and all the variables

-((3x+1)*3)/()+1

We multiply all the terms by the denominator


-((3x+1)*3)+1*()


We calculate terms in parentheses: -((3x+1)*3), so:

(3x+1)*3

We multiply parentheses

9x+3

Back to the equation:

-(9x+3)


We add all the numbers together, and all the variables

-(9x+3)

We get rid of parentheses

-9x-3

Back to the equation:

+(-9x-3)


We get rid of parentheses

2x^2-9x-3=0

a = 2; b = -9; c = -3;
Δ = b2-4ac
Δ = -92-4·2·(-3)
Δ = 105
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{105}}{2*2}=\frac{9-\sqrt{105}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{105}}{2*2}=\frac{9+\sqrt{105}}{4}
$