(7/x)+(3/2x)=10

Solution for (7/x)+(3/2x)=10 equation:

(7/x)+(3/2x)=10

We move all terms to the left:

(7/x)+(3/2x)-(10)=0


Domain of the equation: x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 2x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+7/x)+(+3/2x)-10=0

We get rid of parentheses

7/x+3/2x-10=0

We calculate fractions


14x/2x^2+3x/2x^2-10=0

We multiply all the terms by the denominator


14x+3x-10*2x^2=0

We add all the numbers together, and all the variables

17x-10*2x^2=0

Wy multiply elements

-20x^2+17x=0

a = -20; b = 17; c = 0;
Δ = b2-4ac
Δ = 172-4·(-20)·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-17}{2*-20}=\frac{-34}{-40}
=17/20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+17}{2*-20}=\frac{0}{-40}
=0
$