(y+8/2y-5)=1

Solution for (y+8/2y-5)=1 equation:

(y+8/2y-5)=1

We move all terms to the left:

(y+8/2y-5)-(1)=0


Domain of the equation: 2y-5)!=0

y∈R


We get rid of parentheses

y+8/2y-5-1=0

We multiply all the terms by the denominator


y*2y-5*2y-1*2y+8=0

Wy multiply elements

2y^2-10y-2y+8=0

We add all the numbers together, and all the variables

2y^2-12y+8=0

a = 2; b = -12; c = +8;
Δ = b2-4ac
Δ = -122-4·2·8
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{5}}{2*2}=\frac{12-4\sqrt{5}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{5}}{2*2}=\frac{12+4\sqrt{5}}{4}
$