x-7/4x+5=x+3/x

Solution for x-7/4x+5=x+3/x equation:

x-7/4x+5=x+3/x

We move all terms to the left:

x-7/4x+5-(x+3/x)=0


Domain of the equation: 4x!=0

x!=0/4

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

x-7/4x-(+x+3/x)+5=0

We get rid of parentheses

x-7/4x-x-3/x+5=0

We calculate fractions


x-x+(-7x)/4x^2+(-12x)/4x^2+5=0

We add all the numbers together, and all the variables

(-7x)/4x^2+(-12x)/4x^2+5=0

We multiply all the terms by the denominator


(-7x)+(-12x)+5*4x^2=0

Wy multiply elements

20x^2+(-7x)+(-12x)=0

We get rid of parentheses

20x^2-7x-12x=0

We add all the numbers together, and all the variables

20x^2-19x=0

a = 20; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·20·0
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*20}=\frac{0}{40}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*20}=\frac{38}{40}
=19/20
$