(8y+32)-(y-3)=65/3y

Solution for (8y+32)-(y-3)=65/3y equation:

(8y+32)-(y-3)=65/3y

We move all terms to the left:

(8y+32)-(y-3)-(65/3y)=0


Domain of the equation: 3y)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

(8y+32)-(y-3)-(+65/3y)=0

We get rid of parentheses

8y-y-65/3y+32+3=0

We multiply all the terms by the denominator


8y*3y-y*3y+32*3y+3*3y-65=0

Wy multiply elements

24y^2-3y^2+96y+9y-65=0

We add all the numbers together, and all the variables

21y^2+105y-65=0

a = 21; b = 105; c = -65;
Δ = b2-4ac
Δ = 1052-4·21·(-65)
Δ = 16485
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{-(105)-\sqrt{16485}}{2*21}=\frac{-105-\sqrt{16485}}{42}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(105)+\sqrt{16485}}{2*21}=\frac{-105+\sqrt{16485}}{42}
$