2/5x+17=4/3x-39

Solution for 2/5x+17=4/3x-39 equation:

2/5x+17=4/3x-39

We move all terms to the left:

2/5x+17-(4/3x-39)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 3x-39)!=0

x∈R


We get rid of parentheses

2/5x-4/3x+39+17=0

We calculate fractions


6x/15x^2+(-20x)/15x^2+39+17=0

We add all the numbers together, and all the variables

6x/15x^2+(-20x)/15x^2+56=0

We multiply all the terms by the denominator


6x+(-20x)+56*15x^2=0

Wy multiply elements

840x^2+6x+(-20x)=0

We get rid of parentheses

840x^2+6x-20x=0

We add all the numbers together, and all the variables

840x^2-14x=0

a = 840; b = -14; c = 0;
Δ = b2-4ac
Δ = -142-4·840·0
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14}{2*840}=\frac{0}{1680}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14}{2*840}=\frac{28}{1680}
=1/60
$