(8/3x+2)-(1/x-5)=0

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Solution for (8/3x+2)-(1/x-5)=0 equation: 



(8/3x+2)-(1/x-5)=0



Domain of the equation: 3x+2)!=0

x∈R





Domain of the equation: x-5)!=0

x∈R



We get rid of parentheses

8/3x-1/x+2+5=0

We calculate fractions


8x/3x^2+(-3x)/3x^2+2+5=0

We add all the numbers together, and all the variables

8x/3x^2+(-3x)/3x^2+7=0

We multiply all the terms by the denominator


8x+(-3x)+7*3x^2=0

Wy multiply elements

21x^2+8x+(-3x)=0

We get rid of parentheses

21x^2+8x-3x=0

We add all the numbers together, and all the variables

21x^2+5x=0

a = 21; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·21·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*21}=\frac{-10}{42}
=-5/21
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*21}=\frac{0}{42}
=0
$

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