2y-3/3y+2=2

Solution for 2y-3/3y+2=2 equation:

2y-3/3y+2=2

We move all terms to the left:

2y-3/3y+2-(2)=0


Domain of the equation: 3y!=0

y!=0/3

y!=0

y∈R


We add all the numbers together, and all the variables

2y-3/3y=0

We multiply all the terms by the denominator


2y*3y-3=0

Wy multiply elements

6y^2-3=0

a = 6; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·6·(-3)
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{2}}{2*6}=\frac{0-6\sqrt{2}}{12}
=-\frac{6\sqrt{2}}{12}
=-\frac{\sqrt{2}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{2}}{2*6}=\frac{0+6\sqrt{2}}{12}
=\frac{6\sqrt{2}}{12}
=\frac{\sqrt{2}}{2}
$