2/5x-16=1/3x+10

Solution for 2/5x-16=1/3x+10 equation:

2/5x-16=1/3x+10

We move all terms to the left:

2/5x-16-(1/3x+10)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 3x+10)!=0

x∈R


We get rid of parentheses

2/5x-1/3x-10-16=0

We calculate fractions


6x/15x^2+(-5x)/15x^2-10-16=0

We add all the numbers together, and all the variables

6x/15x^2+(-5x)/15x^2-26=0

We multiply all the terms by the denominator


6x+(-5x)-26*15x^2=0

Wy multiply elements

-390x^2+6x+(-5x)=0

We get rid of parentheses

-390x^2+6x-5x=0

We add all the numbers together, and all the variables

-390x^2+x=0

a = -390; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-390)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-390}=\frac{-2}{-780}
=1/390
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-390}=\frac{0}{-780}
=0
$