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

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

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

We move all terms to the left:

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


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R


We get rid of parentheses

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

We calculate fractions


375x^2/240x^2+(-160x)/240x^2+144x/240x^2-3=0

We multiply all the terms by the denominator


375x^2+(-160x)+144x-3*240x^2=0

We add all the numbers together, and all the variables

375x^2+144x+(-160x)-3*240x^2=0

Wy multiply elements

375x^2-720x^2+144x+(-160x)=0

We get rid of parentheses

375x^2-720x^2+144x-160x=0

We add all the numbers together, and all the variables

-345x^2-16x=0

a = -345; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·(-345)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*-345}=\frac{0}{-690}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*-345}=\frac{32}{-690}
=-16/345
$