(4/y-5)-(7/4y+8)+(3/y-5)=0

Solution for (4/y-5)-(7/4y+8)+(3/y-5)=0 equation:

(4/y-5)-(7/4y+8)+(3/y-5)=0


Domain of the equation: y-5)!=0

y∈R



Domain of the equation: 4y+8)!=0

y∈R


We get rid of parentheses

4/y-7/4y+3/y-5-8-5=0

We calculate fractions


(12y+4)/4y^2+(-7y)/4y^2-5-8-5=0

We add all the numbers together, and all the variables

(12y+4)/4y^2+(-7y)/4y^2-18=0

We multiply all the terms by the denominator


(12y+4)+(-7y)-18*4y^2=0

Wy multiply elements

-72y^2+(12y+4)+(-7y)=0

We get rid of parentheses

-72y^2+12y-7y+4=0

We add all the numbers together, and all the variables

-72y^2+5y+4=0

a = -72; b = 5; c = +4;
Δ = b2-4ac
Δ = 52-4·(-72)·4
Δ = 1177
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{1177}}{2*-72}=\frac{-5-\sqrt{1177}}{-144}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{1177}}{2*-72}=\frac{-5+\sqrt{1177}}{-144}
$