(3/2x)+(8/x)=5

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

(3/2x)+(8/x)=5

We move all terms to the left:

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


Domain of the equation: 2x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


3x/2x^2+16x/2x^2-5=0

We multiply all the terms by the denominator


3x+16x-5*2x^2=0

We add all the numbers together, and all the variables

19x-5*2x^2=0

Wy multiply elements

-10x^2+19x=0

a = -10; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·(-10)·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*-10}=\frac{-38}{-20}
=1+9/10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*-10}=\frac{0}{-20}
=0
$