2y/y-4+2y-5/y-3=25/3

Solution for 2y/y-4+2y-5/y-3=25/3 equation:

2y/y-4+2y-5/y-3=25/3

We move all terms to the left:

2y/y-4+2y-5/y-3-(25/3)=0


Domain of the equation: y!=0

y∈R


We add all the numbers together, and all the variables

2y/y+2y-5/y-4-3-(+25/3)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

2y+2y/y-5/y-7-25/3=0

We calculate fractions


2y+(2y-15)/3y+(-25y)/3y-7=0

We multiply all the terms by the denominator


2y*3y+(2y-15)+(-25y)-7*3y=0

Wy multiply elements

6y^2+(2y-15)+(-25y)-21y=0

We get rid of parentheses

6y^2+2y-25y-21y-15=0

We add all the numbers together, and all the variables

6y^2-44y-15=0

a = 6; b = -44; c = -15;
Δ = b2-4ac
Δ = -442-4·6·(-15)
Δ = 2296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2296}=\sqrt{4*574}=\sqrt{4}*\sqrt{574}=2\sqrt{574}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-44)-2\sqrt{574}}{2*6}=\frac{44-2\sqrt{574}}{12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-44)+2\sqrt{574}}{2*6}=\frac{44+2\sqrt{574}}{12}
$