(12/7x)+(3/2x)=15/14

Solution for (12/7x)+(3/2x)=15/14 equation:

(12/7x)+(3/2x)=15/14

We move all terms to the left:

(12/7x)+(3/2x)-(15/14)=0


Domain of the equation: 7x)!=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

(+12/7x)+(+3/2x)-(+15/14)=0

We get rid of parentheses

12/7x+3/2x-15/14=0

We calculate fractions


(-420x^2)/196x^2+336x/196x^2+294x/196x^2=0

We multiply all the terms by the denominator


(-420x^2)+336x+294x=0

We add all the numbers together, and all the variables

(-420x^2)+630x=0

We get rid of parentheses

-420x^2+630x=0

a = -420; b = 630; c = 0;
Δ = b2-4ac
Δ = 6302-4·(-420)·0
Δ = 396900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{396900}=630$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(630)-630}{2*-420}=\frac{-1260}{-840}
=1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(630)+630}{2*-420}=\frac{0}{-840}
=0
$