2/3y+1=16y+8

Solution for 2/3y+1=16y+8 equation:

2/3y+1=16y+8

We move all terms to the left:

2/3y+1-(16y+8)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R


We get rid of parentheses

2/3y-16y-8+1=0

We multiply all the terms by the denominator


-16y*3y-8*3y+1*3y+2=0

Wy multiply elements

-48y^2-24y+3y+2=0

We add all the numbers together, and all the variables

-48y^2-21y+2=0

a = -48; b = -21; c = +2;
Δ = b2-4ac
Δ = -212-4·(-48)·2
Δ = 825
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:

$\sqrt{\Delta}=\sqrt{825}=\sqrt{25*33}=\sqrt{25}*\sqrt{33}=5\sqrt{33}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-5\sqrt{33}}{2*-48}=\frac{21-5\sqrt{33}}{-96}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+5\sqrt{33}}{2*-48}=\frac{21+5\sqrt{33}}{-96}
$