(7/2x)+(5/6x)=7/4

Solution for (7/2x)+(5/6x)=7/4 equation:

(7/2x)+(5/6x)=7/4

We move all terms to the left:

(7/2x)+(5/6x)-(7/4)=0


Domain of the equation: 2x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 6x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+7/2x)+(+5/6x)-(+7/4)=0

We get rid of parentheses

7/2x+5/6x-7/4=0

We calculate fractions


(-504x^2)/192x^2+672x/192x^2+160x/192x^2=0

We multiply all the terms by the denominator


(-504x^2)+672x+160x=0

We add all the numbers together, and all the variables

(-504x^2)+832x=0

We get rid of parentheses

-504x^2+832x=0

a = -504; b = 832; c = 0;
Δ = b2-4ac
Δ = 8322-4·(-504)·0
Δ = 692224
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{692224}=832$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(832)-832}{2*-504}=\frac{-1664}{-1008}
=1+41/63
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(832)+832}{2*-504}=\frac{0}{-1008}
=0
$