1/x+6/5=40/10x^-5

Solution for 1/x+6/5=40/10x^-5 equation:

1/x+6/5=40/10x^-5

We move all terms to the left:

1/x+6/5-(40/10x^-5)=0


Domain of the equation: x!=0

x∈R



Domain of the equation: 10x^-5)!=0

x∈R


We get rid of parentheses

1/x-40/10x^+5+6/5=0

We calculate fractions


60x^2/250x^2+250x/250x^2+(-1000x)/250x^2+5=0

We multiply all the terms by the denominator


60x^2+250x+(-1000x)+5*250x^2=0

Wy multiply elements

60x^2+1250x^2+250x+(-1000x)=0

We get rid of parentheses

60x^2+1250x^2+250x-1000x=0

We add all the numbers together, and all the variables

1310x^2-750x=0

a = 1310; b = -750; c = 0;
Δ = b2-4ac
Δ = -7502-4·1310·0
Δ = 562500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{562500}=750$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-750)-750}{2*1310}=\frac{0}{2620}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-750)+750}{2*1310}=\frac{1500}{2620}
=75/131
$