3/5x-19=7/2x+25

Solution for 3/5x-19=7/2x+25 equation:

3/5x-19=7/2x+25

We move all terms to the left:

3/5x-19-(7/2x+25)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 2x+25)!=0

x∈R


We get rid of parentheses

3/5x-7/2x-25-19=0

We calculate fractions


6x/10x^2+(-35x)/10x^2-25-19=0

We add all the numbers together, and all the variables

6x/10x^2+(-35x)/10x^2-44=0

We multiply all the terms by the denominator


6x+(-35x)-44*10x^2=0

Wy multiply elements

-440x^2+6x+(-35x)=0

We get rid of parentheses

-440x^2+6x-35x=0

We add all the numbers together, and all the variables

-440x^2-29x=0

a = -440; b = -29; c = 0;
Δ = b2-4ac
Δ = -292-4·(-440)·0
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-29}{2*-440}=\frac{0}{-880}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+29}{2*-440}=\frac{58}{-880}
=-29/440
$