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

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

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

We move all terms to the left:

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


Domain of the equation: 3)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

-2x^2*5+6x*5-16=0

Wy multiply elements

-10x^2+30x-16=0

a = -10; b = 30; c = -16;
Δ = b2-4ac
Δ = 302-4·(-10)·(-16)
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{65}}{2*-10}=\frac{-30-2\sqrt{65}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{65}}{2*-10}=\frac{-30+2\sqrt{65}}{-20}
$